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Physics 1100: Work & Energy Solutions

  1. In the diagram below, calculate the work done if

    (a) F = 15.0 N, θ = 15°, and Δx = 2.50 m,

    (b) F = 25.0 N, θ = 75°, and Δx = 12.0 m,

    (c) F = 10.0 N, θ = 135°, and Δx = 5.50 m,

    For constant forces, work is defined by W = FΔxcos(θ).

    (a) W = 36.2 J
    (b) W = 77.6 J
    (c) W = -38.9 J

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  2. In the diagram below, a rope with tension T = 150 N pulls a 15.0-kg block 3.0 m up an incline (θ = 25.0°). The coefficient of kinetic friction is μk = 0.20. Find the work done by each force acting on the block.

    To find the work done by a force, we need to know the magnitude of the force and the angle it makes with the displacement. To find forces, we draw a FBD and use Newton's Second Law.

    i
    j
    Fx = max Fy = may 
    T - fk - mgsin(θ) = ma N - mgcos(θ) = 0

    The second equation informs us that N = mgcos(θ). We know fk = μkN = μkmgcos(θ).

    Force Force (N)  θ W = FΔxcos(θ)(J) 
    Tension 150 0  450 
    Weight 147.15 θ + π/2 -187
    Normal 133.36 π/2 0
    Friction 26.67 π -80

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  3. A winch lifts a 150 kg crate 3.0 m upwards with an acceleration of 0.50 m/s2. How much work is done by the winch? How much work is done by gravity?

    To find the work done by a force, we need to know the magnitude of the force and the angle it makes with the displacement. To find forces, we draw a FBD and use Newton's Second Law.

    j
    Fy = may
    T - mg = ma

    The work done by the winch is the work done by tension. The work done by gravity is the work done by the object's weight. Since we know m and g, we find T = mg + ma = 1546.5 N. The work done by tension is Wtension = TΔycos(0) = 4.64 × 103 J. The work done by gravity is Wgravity = mgΔycos(π) = -4.41 × 103 J.

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  4. What work does a baseball bat do on a baseball of mass 0.325 kg which has an initial speed forward of 36 m/s and a final speed of 27 m/s backwards. Assume motion is linear and horizontal. The work done by the bat is a non-conservative force.

    Since we are asked for the work done and have a change in speed, we make use of the generalized Work-Energy Theorem. Since the height of the ball does not change, there is only a change in kinetic energy.

    WNC = E = Kf - Ki = ½m[(vf)2 - (v0)2] = ½(0.325kg)[(-27 m/s)2 - (36 m/s)2] = -92.1 J .

    This is the work done on the ball by the bat. It's not a good hit as the ball slowed down. The batter decreased the energy of the ball. Perhaps he was trying for a bunt!

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    5. Work and Energy: Wexternal = ΔEsystem

  5. Consider a 0.50-kg block travelling at 2.0 m/s on a horizontal surface. At the instant shown, there is a rope with a tension of 1.0 N attached while the block travels 1.0 m to the right. In each case:
    a. Determine what object(s) make(s) up the system.
    b. Determine if the system is isolated or if there are external forces are acting on the system.
    c. Determine if there are internal forces to the system.
    d. Find the final speed of the block.

    (i)
    (a)The block is the system
    (b)The tension in string will do work. The normal and the weight do no work since they are at 90° to the motion. The system is not isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = +TL. There is no change in height only a change in speed, so only the kinetic energy changes. Our equation for this case is

    +TL = ½mvf2 − ½mvi2

    Rearranging yields

    vf2 = vi2 + 2TL/m

    Solving, we find vf = [(2 m/s)2 + 2(1.0 N)(1.0 m)/(0.5 kg)]½ = 2.83 m/s.
    (ii)
    (a)The block is the system
    (b)The tension in string will do work. The normal and the weight do no work since they are at 90° to the motion. The system is not isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = +TLcos(60°). There is no change in height only a change in speed, so only the kinetic energy changes. Our equation for this case is

    +TLcos(60°) = ½mvf2 − ½mvi2

    Rearranging yields

    vf2 = vi2 + 2TLcos(60°)/m

    Solving, we find vf = [(2 m/s)2 + 2(1.0 N)(1.0 m)(0.5)/(0.5 kg)]½ = 2.45 m/s.
    (iii)
    (a)The block is the system
    (b)The tension, the normal and the weight do no work since they are at 90° to the motion. The system is isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = 0. There is no change in height only a change in speed, so only the kinetic energy changes. But with no external work being done, the block maintins a constant 2.0 m/s speed.
    (iv)
    (a)The block is the system
    (b)The tension in string will do work. The normal and the weight do no work since they are at 90° to the motion. The system is not isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = −TL. There is no change in height only a change in speed, so only the kinetic energy changes. Our equation for this case is

    −TL = ½mvf2 − ½mvi2

    Rearranging yields

    vf2 = vi2 − 2TL/m

    Solving, we find vf = [(2 m/s)2 − 2(1.0 N)(1.0 m)/(0.5 kg)]½ = 0 m/s.
    (v)
    (a)The block is the system
    (b)The tension in string and kinetic friction will do work though in opposite directions. The normal and the weight do no work since they are at 90° to the motion. The system is not isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = +TL - fkL. Using Newton's Laws, we note fk = μkN = μk mg. There is no change in height only a change in speed, so only the kinetic energy changes. Our equation for this case is

    TL − μkmgL = ½mvf2 − ½mvi2

    Rearranging yields

    vf2 = vi2 + 2[T − μkmg]L/m

    Solving, we find vf = [(2 m/s)2 + 2[1.0 N − (0.15)(0.5 kg)(9.81 m/s2)](1.0 m)/(0.5 kg)]½ = 2.56 m/s.
    (vi)
    (a)The block is the system
    (b)The kinetic friction will do work. The tension, the normal, and the weight do no work since they are at 90° to the motion. The system is not isolated.
    (c)With one block, there are no internal forces.
    (d)For systems we know the following equation is true, WExternal = ΔEsystem. By inspection, we see WExternal = -fkL. Using Newton's Laws, we note fk = μkN = μk mg. There is no change in height only a change in speed, so only the kinetic energy changes. Our equation for this case is

    −μkmgL = ½mvf2 − ½mvi2

    Eliminating, the common factor m and rearranging yields

    vf2 = vi2 − 2μkgL

    Solving, we find vf = [(2 m/s)2 − 2(0.15)(9.81 m/s2)(1.0 m)]½ = 2.63 m/s.

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  6. Consider the blocks travelling at 5.0 m/s on a horizontal surface in the diagrams below. The blocks travels 2.0 m to the right. In each case:
    a. Determine what object(s) make(s) up the system.
    b. Determine if the system is isolated or if there are external forces are acting on the system.
    c. Determine if there are internal forces to the system.
    d. Find the final speed of the block(s)

    (i)
    (a)The two blocks and connecting string are the system
    (b)No one is pulling on the block and there is no friction acting. Since there are no external forces acting other than weight and the normal force, the system is isolated.
    (c)The tension in the connecting string is an internal force.
    (d)For systems we know the following equation is true, WExternal = Efinal − Efinal. However by inspection, we found WExternal = 0 which means Efinal = Efinal. Now the only energy involved is kinetic energy as there is no change in height. This means that the kinetic energy and thus the speed cannot change, vfinal = 3 m/s.
    (ii)
    (a)The two blocks and connecting string are the system
    (b)No one is pulling on the block but there is kinetic friction acting on each block. The system is not isolated.
    (c)The tension in the connecting string is an internal force.
    (d)For systems we know the following equation is true, WExternal = Efinal − Efinal. Here we find WExternal = −f1kL + −f2kL. For each block, using Newton's Laws, we see f1k = μkN1 = μkm1g and f2k = μkN2 = μkm2g. which means Efinal = Efinal. Now the only energy involved is kinetic energy as there is no change in height. This means we can rewrite our starting equation as

    −μkm1gL + −μkm2gL = [½m1v1f2 + ½m2v2f2] − [½m1v1i2 + ½m2v2i2]


    Now m1 + m2 is a common factor. Dividing through by the common factor leaves

    −μkgL = [½v1f2 + ½v2f2] − [½v1i2 + ½v2i2]


    Also the blocked are connected by a string and must have the same speed. That is v2f = v1f = vf and v2i = v1i = vi. So we have the further simplification

    −μkgL = vf2 − vi2


    Solving, we find vf = [(3 m/s2 − (0.15)(9.81 m/s2)(2.0 m)]½ = 2.46 m/s.
    (iii)
    (a)The two blocks and connecting string are the system.
    (b)There is kinetic friction acting but only on the 1.0-kg block. The system is not isolated.
    (c)The tension in the connecting string is an internal force.
    (d)For systems we know the following equation is true, WExternal = Efinal − Efinal. Here we find WExternal = −f1kL. For the back block, using Newton's Laws, we see f1k = μkN1 = μkm1g. Now the only energy involved is kinetic energy as there is no change in height. This means we can rewrite our starting equation as

    −μkm1gL = [½m1v1f2 + ½m2v2f2] − [½m1v1i2 + ½m2v2i2]


    Also the blocks are connected by a string and must have the same speed. That is v2f = v1f = vf and v2i = v1i = vi. So we have the further simplification

    −μkm1gL = ½(m1 + m2)(vf2 − vi2)


    Rearranging to isolate vf2 yields

    vf2 = vi2 − 2μkm1gL /(m1 + m2)

    Solving, we find vf = [(3 m/s)2 − (2)(0.15)(1.0 kg)(9.81 m/s2)(2.0 m)/(1.5 kg)]½ = 2.25 m/s.

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  7. Consider the two blocks travelling at 5.0 m/s on a horizontal surface in the diagrams below. The blocks will one by one slide down the 3.0 m-long, 40° incline to the lower side. The string between the blocks is long enough so that the first block will be down on the lower level before the second block hits the incline. In each case:
    a. Determine what object(s) make(s) up the system.
    b. Determine if the system is isolated or if there are external forces are acting on the system.
    c. Determine if there are internal forces to the system.
    d. Find the final speed of the blocks when just the first block reaches the lower level.
    e. Find the final speed of the blocks when both blocks reach the lower level.

    (a) The two blocks and connecting string are the system.

    (b) No one is pulling on the block and there is no friction acting. Since there are no external forces acting other than weight and the normal force, the system is isolated.

    (c) The tension in the connecting string is an internal force.

    (d) For systems we know the following equation is true, WExternal = ΔEsystem. However by inspection, we found WExternal = 0 which means ΔEsystem = 0.

    The back 1.0-kg block only has a change in speed and therefore only a change in kinetic energy. The front 0.5-kg block has a change in speed and height, so its KE and PE both change. Our equation for this problem is

    {½m1v1f2 - ½m1v1i2} + {½m2v2f2 - ½m2v2i2} + {m2ghfinal − m2ghinitial} = 0

    The blocks are connected by a string and must have the same speed. That is v2f = v1f = vf and v2i = v1i = vi. So we have the further simplification

    {½[m1 + m2]vf2 - ½[m1 + m2]vi2} + m2g[hfinal − hinitial] = 0

    For convenience take hfinal so that hinitial = (3.0 m)sin(50°) = 1.9284 m. We can rearrange the above equation as

    vf2 = vi2 + 2m2g[hfinal − hinitial] / [m1 + m2] = 0

    Solving, we find vf = [(2 m/s)2 + (2)(0.5 kg)(9.81 m/s2)(1.9284 m)/(1.5 kg)]½ = 4.08 m/s.

    (e) Here both block have a change in KE and PE, so the equation is

    {½[m1 + m2]vf2 - ½[m1 + m2]vi2} + [m1 + m2]g[hfinal − hinitial] = 0

    The term [m1 + m2] is a common factor that cancels out leaving

    ½vf2 - ½vi2 + g[hfinal − hinitial] = 0

    We rearrange to get

    vf2 = vi2 + 2g[hfinal − hinitial] = 0

    Solving, we find vf = [(2 m/s)2 + (2)(9.81 m/s2)(1.9284 m)]½ = 6.46 m/s.

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  8. How much work must be done to stop a 2000-kg car travelling at 60 km/h in 15.0 m? What was the average breaking force?

    Since we are asked for the work done and have a change in speed, we make use of the generalized Work-Energy Theorem. Since the height of the car does not change, there is only a change in kinetic energy. First converting the initial velocity into SI

    60 km/h = 60 km/h × (1000 m)/(1 km) × (1 h)/(3600 s) = 16.67 m/s .

    Therefore,

    Wbrake = E = Kf - Ki = ½m[(vf)2-(v0)2] = ½(2000 kg)[(0)2-(16.67 m/s)2] = 2.778105 J .

    Now the force doing this work, fbrake, is related to the work by Wbrake = fbrakexcos(θ). Since the force is slowing the car down, θ = 180°, cos(180°) = -1, and

    fbrake = -Wbrake / Δx = -(2.778 × 105 J )/(15.0 m) = 1.85 × 104 N .

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  9. In the diagram below, determine the speed of the block at each point. Assume no friction. The mass of the block is 10.0 kg.

    Since the problem involves a change in height and speed, we make use of the generalized Work-Energy Theorem,

    WNC = E = Pf - Pi + Kf - Ki = mg(hf - hi) + ½m[(vf)2-(vi)2] .

    Since there is no mention of friction, WNC = 0. Our equation therefore simplifies to

    mg(hf - hi) + ½m[(vf)2-(vi)2] = 0 ,

    or more simply

    mghf + ½m(vf)2 = mghi + ½m(vi)2 .

    We can divide through by m, and since we know hf, hi, and vi, we can rearrange the above to find vf

    (vf)2 = (vi)2 + 2g(hi - hf) .

    For the given values, we find


    		hf (m) vi (m/s) 
    1 15 5
    2 10 11.1
    3 5 14.9
    4 0 17.9

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  10. A 2.0-kg rock is thrown with initial speed of 9.8 m/s at an unknown angle. The speed of the rock at the top of the parabola is 2.1 m/s. How high does it go? Assume no air resistance.

    Since the problem involves a change in height and speed, we make use of the generalized Work-Energy Theorem,

    WNC = E = Pf - Pi + Kf - Ki = mg(hf - hi) + ½m[(vf)2-(vi)2] .

    Since we are told that there is no air resistance, WNC = 0. Our equation therefore simplifies to

    mg(hf - hi) + ½m[(vf)2-(vi)2] = 0 ,

    or more simply

    mghf + ½m(vf)2 = mghi + ½m(vi)2 .

    We can divide through by mg, and since we know hi, vi, and vf, we can rearrange the above to find hf

    hf = [(vi)2 - (vf)2]/2g = [(9.8 m/s)2 - (2.1 m/s)2]/(29.81 m/s2) = 4.67 m .

    The rock reaches 4.67 m up into the air.

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  11. In the diagram below, a 5.00-kg block slides from rest at a height of h1 = 1.75 m down to a horizontal surface where it passes over a 2.00-m rough patch. The rough patch has a coefficient of kinetic friction μk = 0.25. What height, h2, does the block reach on the incline?

    Since the problem involves a change of height and speed, we make use of the Generalized Work-Energy Theorem. Since the block's initial and final speeds are zero, we have

    WNC = E = Uf - Ui = mgh2 - mgh1 . (1)

    The nonconservative force in this problem is friction. To find the work done by friction, we need to know the friction. To find friction, a force, we draw a FBD at the rough surface and use Newton's Second Law.

    i
    j
    Fx = max Fy = may 
    - fk = -ma N - mg = 0 

    The second equation gives N = mg and we know fk = μkN, so fk = μkmg. Therefore, the work done by friction is Wfriction = -fkΔx = -μkmgΔx. Putting this into equation (1) yields

    -μkmgΔx = mgh2 - mgh1 .

    Solving for h2, we find

    h2 = h1 - μkΔx = 1.25 m .

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  12. In the diagram below, a 5.00-kg block slides from rest at a height of h1 = 1.75 m down to a smooth horizontal surface until it encounters a rough incline. The incline has a coefficient of kinetic friction μk = 0.25. What height, h2, does the block reach on the θ = 30.0° incline?

    Since the problem involves a change of height and speed, we make use of the Generalized Work-Energy Theorem. Since the block's initial and final speeds are zero, we have

    WNC = E = Uf - Ui = mgh2 - mgh1 . (1)

    The nonconservative force in this problem is friction. To find the work done by friction, we need to know the friction. To find friction, a force, we draw a FBD at the rough surface and use Newton's Second Law.

    i
    j
    Fx = max Fy = may 
    -fk - mgsin(θ) = -ma N - mgcos(θ) = 0

    The second equation gives N = mgcos(θ) and we know fk = μkN, so fk = μkmgcos(θ). Therefore, the work done by friction is Wfriction = -fkΔx = -μkmgcos(θ)Δx. Putting this into equation (1) yields

    -μkmgcos(θ)Δx = mgh2 - mgh1 .

    A little trigonometry shows that Δx is related to h2 by Δx = h2 / sin(θ). Putting this into the above equation yields

    -μkcos(θ)[ h2 / sin(θ)] = h2 - h1 .

    Solving for h2, we find

    h2 = h1 / [1 + μk/tan(θ)] = 1.22 m .

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  13. In the figure below, a block of mass 5.0 kg starts at point A with a speed of 15.0 m/s on a flat frictionless surface. At point B, it encounters an incline with coefficient of kinetic friction μk = 0.15. The block makes it up the incline to a second flat frictionless surface. What is the work done by friction? What is the velocity of the block at point C? The incline is 2.2 m long at an angle θ = 15°.

    The problem involves a change in height and speed, so we apply the generalized Work-Energy Theorem.

    WNC = E = (Kf - Ki) + (Uf - Ui) = ½m(vC)2 - ½m(vA)2 + mgh . (1)

    Here the nonconservative force is friction, so WNC = Wf. To find friction, a force, we draw a FBD and use Newton's Second Law.

    i
    j
    Fx = max Fy = may 
    -fk - mgsin(θ) = -ma N - mgcos(θ) = 0

    The second equation gives N = mgcos(θ) and we know fk = μkN, so fk = μkmgcos(θ). Therefore, the work done by friction is

    Wf = -fkΔx = -μkmgcos(θ)Δx = -(.15)(5 kg)(9.81 m/s2)cos(15°)(2.2 m) = -15.635 J.

    Note from the diagram, that the height h is related to the length of the incline by h = Δxsin(θ). Putting both results into equation (1) yields

    Wf = ½m(vC)2 - ½m(vA)2 + mg[Δxsin(θ)] .

    Solving for vC yields

    vC = [2Wf/m - 2gΔxsin(θ) + (vA)2]½ = 14.4 m/s .

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  14. Two blocks are connected by a string hung over a frictionless massless pulley. Block A has mass MA and block B has mass MB. Initially the blocks are held at rest before being allowed to move. How fast will block B be moving when it has risen a distance h?

    15. Again we have a change in height and speed, so we apply the Work-Energy Theorem

    WNC = (Kf - Ki) + (Uf - Ui).

    We are told that there is no friction so WNC = 0.

    The difference between this and earlier problems is that we are dealing with two objects.  For each object there is an external force the tension T in the string.  However the work done by the tension in each case is equal, since the distance each block moves is the same, but opposite. (Check this!)  So for the system, energy is transferred from one block to the other.  We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn. 

    0 = [½MBvf2 + MBgh] + [½MAvf2 − MAgh] 

    Note that the two blocks are connected by a string so the final speed of each is the same. Also if block B moves up h block A drops h. Thus our equation becomes

    0 = ½ (MA + MB)vf2 − (MA − MB)gh.

    When we solve this, we find

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  15. Two blocks are connected by a string hung over a frictionless massless pulley. Block A has mass MA and is on a table top. Block B has mass MB and is hanging in the air. Initially the blocks are held at rest. The coefficients of friction between block A and the tabletop are μS and μK.
    (a) B is allowed to fall. How fast will block B be moving when it has fallen distance h?
    (b) Block A is pulled to the left by a horizontal force F for a distance L. How fast will block B be moving?

    16. (a) Again we have a change in height and speed, so we apply the Work-Energy Theorem

    WNC = (Kf - Ki) + (Uf - Ui).

    We are told that there is friction so we need to determine WNC = Wfriction. Friction does negative work, takes energy out of the system, since it is opposite to the movement of block A. To find friction, a force, we draw a FBD of block A and use Newton's Second Law.

    i

    j

    ΣFx = max

    ΣFy = may

    T - fk = MAa

    N - MAg = 0

    The second equation gives N = MAg and we know fk = μkN, so fk = μkMAg. Therefore, the work done by friction is Wfriction = -fkΔx = -μkMAgh since block A will move as far as block B will drop.

    For the pair of blocks, the tension T in the string, is internal and does not net work. So for the system, energy is transferred from one block to the other.  We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

    -μkMAgh  = ½MAvf2 + [½MBvf2 − MAgh]

    Note that the two blocks are connected by a string so the final speed of each is the same. Thus our equation becomes

    MBgh − μkMAgh = ½(MA + MB)vf2.

    When we solve this, we find

    (b) Again we have a change in height and speed, so we apply the Work-Energy Theorem

    WNC = (Kf - Ki) + (Uf - Ui).

    We are told that there is friction, and the work done by friction is still Wfriction = -fkΔx = -μkMAgL since block A moves L not h. Because of the string block B rises L and both blocks will have the same speed. However there is an extra external force F which in the same direction as the motion of block A. It does positive work adding to the energy of the system.

    We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

    FL − μkMAgh = ½MAvf2 + [½MBvf2 + MAgL]

    Thus our equation becomes

    FL − MBgh − μkMAgh = ½(MA + MB)vf2 .

    When we solve this, we find

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  16. Two blocks are connected by a sting slung over a pulley as shown in the diagram below. The hanging block is allowed to drop.  How fast will it be moving when it hits the ground? The block on the incline has mass MA = 2.50 kg. The hanging block has mass MB = 1.50 kg. The incline makes and angle θ = 30° with horizontal.  Ignore friction.

    Again we have a change in height and speed, so we apply the Work-Energy Theorem

    WNC = (Kf - Ki) + (Uf - Ui).

    We are told to ignore friction so WNC = 0.

    The difference between this and earlier problems is that we are dealing with two objects. For each object there is an external force the tension T in the string. However the work done by the tension in each case is equal, since the distance each block moves is the same, but opposite. (Check this!) So for the system, energy is transferred from one block to the other. We solve the problem by applying the right hand side of the Work-Energy Theorem to each block in turn.

    0  = (½MBVBf2 - 0) + (MBg(0) - MBg(1.0m))

    		+ (½MAVAf2 - 0) + (MAg(hAf - hAi))

    Now the two blocks are connected by a string so the final speed of each is the same, VBf = VAf = Vf. Next the block moves 1.0 m up the 30° degree incline, so hAf - hAi = (1.0 m)sin(30°). Thus our equation becomes

    0 = ½MBVf2 + ½MAVf2 - MBg(1.0m) + MAg(1.0m)sin(30°) .

    When we solve this we find

    Vf = {2(9.81)[(1.50)(1.0m) - (2.50)(1.0m)sin(30°)]/(1.50 + 2.50)}½ = 1.338 m/s .

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  17. In the diagram below, what is the minimum height that the skier must start from to successfully make it around the loop. Assume (a) no friction, and (b) that friction does -3.0 × 103 J of work on the skier. The radius is 5.00 m and the skier has mass 65.0 kg.

    The problem involves a change of height and speed, so that suggests that we use the generalized Work-Energy Equation. However, the skier also travels in a circle, which suggests a centripetal acceleration problem. Centripetal acceleration problems are solved by drawing a free-body diagram (FBD) and applying Newton's Second Law. Let's do this first.

    At the top of the inside of the loop, the centripetal acceleration acts straight down as does the normal force and the weight.

    j
    Fy = may
    -N - mg = -m(vf)2/r

    The skier will lose contact with the inside of the loop when N goes to zero. This fact and our equation, let's us find a minimum value of vf,

    vf = [gr]½ = [(9.81 m/s2)(5.00 m)]½ = 7.004 m/s .

    Now we consider the work energy portion of the problem.

    The Work-Energy formula may be rewritten as

    mghf + ½m(vf)2 = mghi + ½m(vi)2 .

    We know vi = 0, we see from the diagram that hf = 2r, and vf = [gr]½ from our earlier work, so we rearrange the above equation to find hi

    hi = hf + ½(vf)2/g = 2r + ½r = (5/2)r = 12.5 m .

    If the trip is frictionless, the hill needs to be at least 12.5-m tall if the skier is to make it around the loop safely.

    Since there are non-conservative forces, the generalized Work-Energy equation for this case is

    WNC = [mghf + ½m(vf)2] - [mghi + ½m(vi)2] .

    We are told WNC = -3000 J, so we rearrange the equation to find that hi is,

    hi = {[mghf + ½m(vf)2] - WNC }/mg = (5/2)r - WNC/mg .

    Using the given data,

    hi = 12.5 m - (-3000 J)/(65.0 kg)(9.81 m/s2) = 17.2 m .

    With this much friction, the hill needs to be at least 17.2-m tall if the skier is to make it around the loop safely.

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  18. Tarzan, Lord of Apes, is swinging through the jungle. In the diagram below, Tarzan is standing at point A on a tree branch h1 = 22.0 m above the floor of the jungle. Tarzan is holding one end of a vine which is attached to a branch on a second tree. The vine is L = 21.0 m long. When Tarzan swings on the vine, his path is in an arc of a circle. At the bottom of his swing he is at point B, 13.0 m above the ground . Ignore Tarzan's height. Tarzan has a mass of 90.0 kg. The vine does not stretch and has negligible mass.
    (a) Why does the tension in the vine do no work?
    (b) What will be his speed at point B?
    (c)What will be the tension in the rope at point B?

    The problem involves a change in height and speed, so we apply the generalized Work-Energy Theorem.

    WNC = E = (Kf - Ki) + (Uf - Ui) = ½m(vB)2 - mgh . (1)
    (a) Here the only possible nonconservative force is friction, so WNC = WT. The definition of work is W = Fxcos, but in this problem the tension is along a radius and is thus always at 90 to the displacement. As a result, WT = 0. Thus we have
    0 = ½m(vB)2 - mgh .

    (b) To find the speed at point B, we need to know h, the distance Tarzan dropped. Examining the question, we see that h = h1 - h2 = 22.0 m - 13.0 m = 9.0 m. Rearranging our equation, we find

    vB = [2gh]½ = [2(9.81 m/s2)(9 m)]½ = 13.29 m/s .

    (c) Tension is a force. To find a force we need to draw a FBD and apply Newton's Second Law. Since Tarzan is swinging in a circle, we are dealing with centripetal acceleration.

    j
    Fy = may
    T - mg = mv2/L

    Solving for T,

    T = mg + mv2/L = (90.0 kg)[ 9.81 m/s2 + (13.288 m/s)2/(21.0 m)] = 1.64 103 N.

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  19. A 0.200-kg block slides down the track and horizontally off a table as shown in the diagram below.
    (a) Assuming that friction is negligible, how far from the table does the block land?
    (b) The block only land 1.20 m away. How much work was done by friction and other non-conservative forces?

    The first part of the problem involves a change in height and speed, so we can use the Work-Energy Theorem there. When the block leaves the surface it becomes a projectile.

    (a) Applying the Work-Energy Theorem and assuming that the initial velocity of the block is zero.

    WNC = ΔE = (Kf - Ki) + (Uf - Ui) = ½m(vB)2 - 0 + mg(hf - hi) .

    The mass m cancels out and we find

    vB = [-2g(hf - hi)]½ = [-2(9.81)(1.0 - 1.5)]½ = 3.3121 m/s .

    Now this velocity is the initial velocity for the projectile.

    i j
    v0x = 3.3121 m/s v0y = 0 m/s (horizontal flight)
    ax = 0 m/s2 ay = -9.81 m/s2
    Δx = ? Δy = -1.0 m
    ----- t (common) -----

    From the j information we can find the time that the block is in the air using Δy = v0yt + ½ayt2. This becomes -1.0 m = ½(-9.81 m/s2)t2 or t = ±0.4515 s. We need the positive, forward in time, solution. We then find Δx using

    Δx = v0xt + ½axt2 = (3.3121 m/s)(0.4515 s) = 1.4955 m.

    The block lands 1.50 m from the edge of the table.

    (b) If the block only lands 1.20 m away, then is velocity must have been v0x = (1.20 m)/(0.4515 s) = 2.6578 m/s .

    This is also the velocity at the bottom of the slide. To find WNC we again use the Work-Energy Theorem.

    WNC = (Kf - Ki) + (Uf - Ui) = ½m(v0x)2 - 0 + mg(hf - hi) .

    So the work done by non-conservative forces is

    WNC = ½(0.200)(2.6578)2 + (0.200)g(1.0 - 1.5) = -0.2746 J.
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  20. In the diagram below, the spring has a force constant of 5000 N/m, the block has a mass of 6.20 kg, and the height h of the hill is 5.25 m. Determine the compression of the spring such that the block just makes it to the top of the hill. Assume that there are no non-conservative forces involved.

    Since the problem involves a change is height and has a spring, we make use of the Generalized Work-Energy Theorem. Since the initial and final speeds are zero,

    Wext = ΔE = Ugrav f - Uspring i = mgh - ½Kx2 .

    There are no external forces so Wext = 0. Getting x by itself yields

    x = [2mgh / K]½ = 0.357 m .

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  21. A block of mass m is connected by a string of negligible mass to a spring with spring constant K which is in turn fixed to a wall. The spring is horizontal and the string is hung over a pulley such that the mass hangs vertically. The pulley is massless. As shown in the diagram below, the spring is initially in its equilibrium position and the system is not moving.
    (a) Use energy methods, to determine the speed v of the block after it has fallen a distance h. Express your answer in terms of g, m, K, and h.
    (b) The block will oscillate between its initial height and its lowest point. At its lowest point, it turns around. Use your answer to part (a) to find where it turns around.

    (a) The problem involves a change in height and speed and has a spring, so we would apply the generalized Work-Energy Theorem even if not directed to do,

    Wext = ΔE = (Kf − Ki) + (Uf − Ui) ,            (1)

    where K is the sum of all the linear kinetic energies of each object, and U is the sum of the spring and gravitational potential energies. Since there is no kinetic friction acting on the system, Wext = 0.

    Examining the problem object by object we see that the spring stretches, so there is an increase in spring potential energy. The pulley is massless and can be ignored as it can have no kinetic energy if it has no mass. The block drops, so there is a decrease in its gravitational potential energy. As well, as the block drop, it increases its kinetic energy. Equation (1) for this problem is thus

    0 = ½Kx2 − mgh + ½mv2 .

    Since the spring is connected to the block, the spring stretches as much as the block drops, so x = h. Substituting this relation back into our equation yields,

    0 = ½Kh2 − mgh + ½mv2 .

    Collecting the terms with v, and solving for v yields

    v = [(2mgh − kh2) / m]½.            (2)

    (b) Recall from our discussions on kinematics that an object turns around when its velocity is zero. Setting equation (2) to zero

    [(2mgh − kh2) / m)]½ = 0 ,

    we see that the numerator is zero when

    2mgh − kh2 = 0 .

    Solving this for h reveals that the object turns around when h = 2mg/k or when h = 0 which means that the block oscillates between these two heights.

    22. [Return to Top of Page]

  22. At point A in the figure shown below, a spring (spring constant k = 1000 N/m) is compressed 50.0 cm by a 2.00 kg block. When released the block travels over the frictionless track until it is launched into the air at point B. It lands at point C. The inclined part of the track makes an angle of θ = 55.0° with the horizontal and point B is a height h = 4.50 m above the ground. How far horizontally is point C from point B?

    The problem involves a change in height and speed and has a spring, so we apply the generalized Work-Energy Theorem., Wext = ΔE.

    There is no friction or air resistance, so Wext = 0. The spring is compressed initially, so it loses spring potential energy. The block increases kinetic energy and gains gravitational potential energy. Our equation is thus

    0 = −½kx2 + ½mv2 + mgh .

    We can use this to find the speed of the block at launch

    v = [kx2/m − 2gh]½ = [(1000)(0.5)2/2 − 2(9.81)(4.5)]½ = 6.0589 m/s .

    Now the block is a projectile. To solve a projectile problem we break the motion into its x and y components and apply our kinematics equations.

    i
    j
    v0x = vcos(55°) = 3.47523 m/s v0y = vsin(55°) = 4.96314 m/s
    Δx = ? Δy = −4.50 m
    ax = 0 ay = −9.81 m/s2
    t = ? t = ?

    We have enough information in the y column to find t using Δy = v0yt + ½at2 ,

    −4.50 = 4.96314t − 4.905t2 .

    Using the quadratic equation, the solutions are t = −0.5773 s and t = 1.5892 s. We want the positive, or forward in time, solution. Hence the horizontal distance traveled by the block is

    Δx = v0xt = 3.47523 × 1.5892 = 5.52 m .

    Point C is therefore 5.52 m from B.

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  23. A block of mass M on a flat table is connected by a string of negligible mass to a vertical spring with spring constant K which is fixed to the floor. The string goes over a massless pulley. As shown in the diagram below, the spring is initially in its equilibrium position and the system is not moving. A person pulls the block with force F through a distance L. Determine the speed v of the block after it has moved distance L.The tabletop is frictionless.

    The problem involves changes in height, speed, and rotation, so we would apply the generalized Work-Energy Theorem even if not directed to do so, 

    Wext = ΔE, (1)

    where E is the sum of all the mechanical energies of each object. If the system consists of the spring, string, pulley, block and the earth, then F is an external force acting on the system and Wext = FL.

    Next consider the change in energy of each object. The spring stretches as so increases its potential energy. The pulley can be ignored as it is massless. The block moves from rest so it increases its linear kinetic energy. Thus equation (1) becomes

    FL = ½Kx2 + ½Mv2.

    Since the block and spring are connected by the same string, the spring has stretched x = L. Substituting this back yields

    FL = ½KL2 + ½Mv2.

    Taking the term with v to one side yields

    ½Mv2 = FL − ½KL2 .

    Solving for v yields,

    v = [(2FL − KL2) / M]½ .

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    24. Power

  24. A 50-hp engine is used to lift heavy loads at a worksite. It is used to lift a load of bricks weighing 2000 N to the top of a new building 35.0 m above ground. How long does it take for the load to get to the top?

    We are given the power of the engine

    P = 50 hp × (740 W/hp) = 37,000 W.

    Power is defined as work done per given time, P = W/t. The time t is what we are asked for. Work done is force times distance, here W = 2000 N × 35 m = 70,000 J.

    So the time needed is

    t = W/P = 70,000 J / 37,000 W = 1.89 seconds.

    Note however that rated power is seldom the same as the actual power that does useful work.

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  25. A 5.0 MW generating station is situated at a 22-m high dam. The energy used to generate the electricity comes from the loss in potential energy of the water as it falls the height of the dam. What is the minimum amount of water going through the dam every day?

    We are given the power (50,000,000 W) and power is defined as work done per given time, P = W/t. The time t we are given is one day. We are told that the work done equals the loss in potential energy of the water falling from the top of the dam, so W = mgh where h = 22.0 m. Thus the amount of water, i.e. its mass, is found from P = mgh / t or

    m = Pt/gh = (5 × 106)(1 d × 24 h/d × 3600 s/h) / (9.81)(22.0) = 2.00 × 109 kg.

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  26. What power is required to pull a 5.0-kg block at a steady speed of 1.25 m/s? The coefficient of friction is 0.30.

    The power required to move the block at constant speed is P = Fv. We are given v, the speed of the block. To get F, a force, we draw a FBD and apply Newton's Second Law,

    i 
    j
    Fx = max Fy = may 
    F - fk = 0 N - mg = 0 

    The second equation gives N = mg and we know fk = μkN, so fk = μkmg. Therefore, the applied force is F = μkmg. Thus the power is

    P = μkmgv = (0.3)(5 kg)(9.81 m/s2)(1.25 m/s) = 18.4 Watts.

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  27. A 7000-W engine is propelling a speedboat at 30 km/h. What force is the engine exerting on the speedboat? What force and how much power is water resistance exerting on the speedboat?

    First we convert the velocity to SI units,

    30 km/h × (1000 m)/km × (1 h)/(3600 s) = 8.333 m/s .

    We know P = Fv, so

    F = P/v = 7000 W / 8.333 m/s = 840 N .

    By Newton's Third Law, the water is exerting 840 N in the reverse direction. It is also removing 7000 W of power which is going into increasing the kinetic energy of the water.

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