Answers - Work and Energy

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If you do not raise the object or increase its velocity, there is no increase in the object's potential energy or in its kinetic energy. The physicists' definition of work demands that when you do work on an object its energy increases. For the situation in Fig. 1, you are exerting a force through a distance, but you do no work. For this reason, we define work W as the force F times the displacement s times the angle between F and s or F,s. In Fig. 1, F,s = 90⁰ and cos 90⁰ = 0. With this definition of work, we correctly predict that no work is done moving an object horizontally without increasing its speed. You may tire while moving the object because your tensed muscles are continually contracting and relaxing in minute movements. You, however, have done no work on the object as defined in a physical sense.

4.

The force that produces a centripetal acceleration is always in toward the center or perpendicular to the path. The angle between the tension and the displacement is 90⁰. Since cos 90⁰ = 0, no work is done by the tension in the string. Also note that the potential energy and the kinetic energy of the object do not change.

5.

(a) The angle between F and s is 0^o.
     The work done by F = Fs cos 0^o = 12 N(4 m)(1) = 48 J.

(b) The angle between F_N and s is 90^o.
     The work done by F_N = F_Ns cos 90^o = F_Ns(0) = 0.

(c) The angle between mg and s is -90^o.
     The work done by mg = mgs cos -90^o = mgs(0) = 0.

(d) To find the frictional force f,  notice that (F_net)_y = ma_y = m(0) = 0.
     F_N - mg = 0  and  F_N = mg.
     f = µ_kF_N = 0.50(mg) = 0.50(20 N) = 10 N.
     The angle between f and s is 180^o.
     The work done by f = fs cos 180^o = 10 N(4 m)(-1) = - 40 J.

(e) The work done by the net force = 48 J - 40 J = 8 J,  or
     (F_net)_x = 12 N - 10 N = 2 N.
     Work done by net force = 2 N(4 m)cos 0⁰ = 8 J.

6.

For the raindrop to fall with a constant speed, the net force acting on the drop must equal 0. Thus mg - f = 0 or mg = f, where f is the frictional force of the air on the drop and I have taken down as positive.



Work = Force (displacement) F,s. The gravitational force mg is in the same direction as the displacement s. Work by gravitational force = mgs cos 0⁰ =
mgs(1) = 3.35 x 10^-5 kg(10 m/s²)(100 m)(1) = 3.35 x 10^-2 J. The frictional force of the air is opposite to the direction of the displacement s. Work by frictional force of air = fs cos 180⁰ = -3.35 x 10^-2 J.

7.

Kinetic energy K = 1/2 mv² = 1/2(52 kg)(14.0 m/s)² = 51 x 10² J.

8.

(a) In the figure, we take the Y-axis perpendicular to the incline.

(Fnet)y = may
FN - mg cos 370 = m(0)
FN = mg cos 370 = (2.0 kg)(10 m/s2)(0.8) =16N
f = µkFN = 0.50(16 N) = 8.0 N

(b) Work done by F = Fs cos F,s = 30N(4m)(cos 0⁰) = 30 N(4.0 m)(1) =120 J.

(c) Work done by F_N = F_Ns cos 90⁰ = F_Ns(0) = 0.

(d) Angle between mg and s = -(90 + 37)⁰. Work done by mg = mgs cos mg,s   = 2.0 kg(10 m/s²)(4.0 m)cos {-127⁰)} = 2.0 kg(10 m/s²)(4.0 m) (-0.6) = -48 J.

(e) Work done by friction = f(s)cos 180⁰ = 8.0 N(4.0 m)(-1) = -32 J.

(f) Work by net force =120 J - 32 J + 0 - 48 J = 40 J.

(g) By work-energy theorem, work done by net force = 40 J = ΔK
   = 1/2 mv_f² - 1/2 mv_i² = 1/2(2kg)v_f² - 0.
   40 J = 40 N-m = 40 (kg-m/s²)-m = 40 kg-m²/s² = (kg)v_f².
   v_f = √40 m/s = 2√10 m/s.

9.

(a)	For the Y-axis, (Fnet)y = may
	From the figure,
	FN - F sin 370 - mg = 0
	FN = F sin 370 + mg
	FN = (30 N)(0.6) + 20 N = 38 N
	f = µkFN = (0.50)(38 N) = 19 N

(b) Work by f = fs cos f, s = (19 N)(4.0 m)(cos 180^o) = -76 J

(c) Work done by F = Fs cos F,s = 30 N(4.0 m)(cos 37^o) =120 J(0.8) = 96 J

(d) Work done by F_N = F_Ns cos F_N,s = F_Ns cos 90^o = F_Ns (0) = 0

(e) Work done by mg = mgs cos mg, s = mgs cos (-90^o) = mgs (0) = 0

(f) Net work done = Work by f + Work by F + Work by F_N + Work by mg
                             =     -76 J    +      96 J      +          0        +       0       =   20 J

(g) By work energy theorem,
Work done by net force = Δ kinetic energy = K_f - K_i = 1/2 mv_f² - 1/2 mv_i²
20 J = 1/2 (2 kg)v_f² - 0
v_f = (20 J/kg)^1/2 = (20 N-m/kg)^1/2 = {20 (kg-m/s²)(m/kg)}^1/2 = √20 m/s
= 2√5 m/s.

10.

Work = force (displacement) cos F,s.

(a) The angle between the force exerted by the person and the displacement is 0⁰.  Work by person =10 N(1.0 m) = 10 J.

(b) The gravitational force = mg = 1.0 kg(10 m/s²) = 10 N. To lift the object with a constant velocity the force of the person must be equal in magnitude to the gravitational force. The angle between mg and s is 180⁰. Work done by gravity = 10 N(1.0 m)cos 180⁰ = -10 J.

(c) Increase in potential energy = mgh, where h is the height to which the object is raised = 10 J. The increase in the potential energy equals the work done by the person in raising the object with a constant velocity or it equals the negative of the work done by the gravitational force.

(d) The increase in kinetic energy is zero because the work done by the net force equals zero.

11.

(a) Now work done by person = 12 N(1.0 m) = 12 J.

(b) Work done by gravitational force still equals -10 J.

(c) Increase in potential energy equals the negative of the work done by the gravitational force = 10 J.

(d) Work done by the net force = 12 J - 10 J = 2 J = increase in kinetic energy.

12.

(a) Work done by friction = fs cos 180^o = 72 N(3.0 m)(-1) = -216 J.

(b) Work by net force = work done by friction = change in kinetic energy
-216 J = 1/2 (12 kg)v²_3.0m - 1/2(12 kg)(10 m/s)²
-216 J + 600 J= 1/2(12 kg)v²_3.0m.
V_3.0m = 8.0 m/s.

(c) Work done by friction = 72 N(s)(-1) = change in kinetic energy
= 0 - 600 J.    s = 8.3 m

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14.

Take U_f at the bottom of the incline = 0.
Work by friction = fs cos 180^o =(U_f + K_f) - (U_i + K_i).
f(125 m)(-1) = {0 +(1/2)(1.0 kg)(25 m/s)²} - {1.0 kg)(10 m/s²)(62.5 m) + 0}
      -125 m f = 312.5 J - 625 J = -312.5 J.     f = 2.5 N

15.

The total energy E = U + K or K = E - U
  = {-21.7 - (-43.4)}10^-19 J. K = 21.7 x 10^-19 J.

While the total and potential energy are negative, classically the kinetic energy must always be positive.

16.

(a) acceleration a = Δv/ Δt = (0.20 - 0)m/s/0.50 s = 0.40 m/s².
     F = ma = 2000 kg(0.40 m/s²) = 800 N.

(b) K = 1/2(2000 kg)(0.20 m/s)² = 40 J.

(c) Work done by net force = change in kinetic energy = 40 J.

(d) Work = Fs = 800 N s = 40 J.     s = 40 J/800 N = 0.05 m.

17.

(a)   i ^. i = (1)(1) cos 0⁰ = 1

(b)   i ^. j = j ^. i = (1)(1) cos 90⁰ = 0

(c)   j ^. j = (1)(1) cos 0⁰ = 1

(d)   W = F ^. s = 20 N j ^. 4 m i = 80 J(j ^. i) = 0

(e)   W = F ^. s = (3i + 4j)N ^. (2i - 2j)m = 6J(i ^. i) - 6J(i ^. j) + 8J(j ^. i) - 8J(j ^. j)
    = 6 J - 0 + 0 - 8 J = -2 J

18.

Since the incline is frictionless and no other nonconservative force acts on the object, energy is conserved. Take the initial point i at the top of the incline and the final point f at the bottom of the incline. Let U_f = 0. At the initial point the potential energy is mgh, where h is the vertical height above the bottom of the incline. The object being released from the top of the incline means that its initial velocity v_i is zero so that K_i = 1/2 mv_i² = 0.
From conservation of energy,

Ui	+	Ki	=	Uf	+	Kf
2.0 kg(10 m/s2)(3 m)	+	0	=	0	+	1/2(2.0 kg) vf2
60m2/s2			=	vf2
or vf = (60)1/2 m/s = 7.7 m/s

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20.

(a) From Fig. for #20, at y = 0.750 m, U = 12.0 J.

(b) At all times, E = 16.0 J = U + K = U(0.750m) + K(0.750 m)
   = 12.0 J + K(0.750 m) or K(0.750m) = 4.0 J.

(c) U = mgh or at 0.750 m, 12.0 J = m(10 m/s²)(0.75 m) or m = 1.6 kg.

(d) K = 1/2 mv² = 1/2(1.6kg)v² = 4.0 J or v = √ 5 m/s.

21.

(a) We use conservation of energy to solve the problem. We take the gravitational potential energy equal to 0 at the bottom of the loop-the-loop (Fig. for #21). The block is released at the initial position i so the kinetic energy there is zero. The block will make it around the loop-the-loop if it can stay on the loop as it gets to the top of the loop. For this reason we take the top of the loop as the final position f.

At the top, the acceleration is in toward the center.

For the minimum height h, we want the minimum velocity v_f, so we set the normal force F_N = 0.  Then mg = mv_f²/R or:

Substituting Eq. 2 into Eq. 1: mgh = 5/2 mgR or h = 5R/2.

(b) At P, the potential energy = mgR.

At P, the normal force produces the centripetal acceleration into the center of the circle.  At P, mg is down.

22.

We choose different potential energy positions for the potential energy equal to zero for m₁ and m₂. Notice that when m₂ moves down, m₁ moves up the incline. Since we always take the lowest position for the zero of potential energy, the position of m₁ in Fig. 5 (in the statement of the problem) is its lowest position and we take the potential energy of m₁ equal to zero there. The position of m₂ in Fig. 5 is its highest position. We take its zero potential at a point 2.5 m below where it is in Fig. 5, or its dashed position in Fig. 5'.

The work done by friction = f(d) cos f,d

In the final position f₂, the potential energy (U₂)_f of m₂ is zero, but in the final position f₁, the potential energy for m₁ is m₁gh₁, where h₁ = the vertical height above the initial position. Since it goes up the plane 2.5 m and the plane is inclined at 37⁰, the vertical height h₁ = 2.5 m sin 37⁰ = 2.5 m (3/5) = 1.5 m.
(U₁)_f = (m₁g)h₁ = (20 N)1.5 m = 30 J. For the system U_f = (U₁)_f + (U₂)_f = 30 J + 0 = 30 J. Since both blocks move together, their final speed v_f is the same. The final kinetic energy of the system K_f = (1/2)(m₁ + m₂)v_f² = 1/2(6.0 kg) v_f². The initial potential energy (U₁)_i of m₁ at i₁ is zero. The initial potential energy of m₂ at i₂ is (U₂)_i = m₂g(2.5 m) = (4.0 kg)(10 m/s²)(2.5 m) = 100 J. For the system, U_i = 100 J. Both blocks were initially at rest. For the system K_i = 0.

Again,

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