CHAPTER 5

WORK, ENERGY, AND POWER

# C LEVEL PROBLEMS

1)      A 900-N crate rests on the floor. How much work is required to move it at a constant speed (a) 6.0-m along the floor against a frictional force of 180 N, and (b) 6.0-m vertically?

SOLUTION:

(a)  Because there is no acceleration, the horizontal applied force must have the same magnitude as the friction force. Thus,

W = FDx = (180 N)(6.0 m) =1.1 ´ 103 J.

(b)  Because there is no acceleration, the vertical applied force must have the same magnitude as the weight. Thus,

W = FDy = mgDy = (900 N)(6.0 m) =5.4 ´ 103 J.

2)      At room temperature, an oxygen molecule, with a mass of 5.31 x 10-26 kg, typically has a kinetic energy of about 6.21 x 10-21 J. How fast is it moving?

SOLUTION:

We find the speed from

ke = ½ mv2;

6.21 ´ 10–21 J = ½ (5.31 ´ 10–26 kg)v2, which gives v =484 m/s.

3)      How much work is required to stop an electron (m = 9.11 x 10-31 kg) which is moving with a speed of 1.90 x 106 m/s?

SOLUTION:

The work done on the electron decreases its kinetic energy:

W = Dke = ½mv2 – ½mv02 = 0 – ½ (9.11 ´ 10–31 kg)(1.90 ´ 106 m/s)2 = – 1.64 ´ 10–18 J.

4)      A spring has a spring constant, k, of 440 N/m. How much must this spring be stretched in order to store 25 J of potential energy?

SOLUTION:

The potential energy of the spring is zero when the spring is not stretched (x= 0). For the stored potential energy, we have

pe = ½kxf2– 0;

25 J = ½ (440 N/m) xf2 – 0, which gives xf=0.34 m.

5)      By how much does the potential energy of a 64-kg pole-vaulter change if his center of mass rises about 4.0-m during the jump?

SOLUTION:

For the potential energy change we have

Dpe = mg Dy = (64 kg)(9.80 m/s2)(4.0 m) =2.5 ´ 103 J.

6)      A novice skier, starting from rest, slides down a frictionless 35.00 incline whose vertical height is 125 m.How fast is she going when she reaches the bottom?

SOLUTION:

We choose the potential energy to be zero at the bottom (y = 0). Because there is no friction and the normal force does no work, energy is conserved, so we have

ekepekepe;

½mvi2 + mgyi = ½mvf2 + mgy;

½m(0)2 + m(9.80 m/s2)(125 m) = ½mvf2 + m(9.80 m/s2)(0), which gives vf =49.5 m/s.

This is 180 km/h! It is a good thing there is friction on the ski slopes.

7)      How long will it take a 1750-W motor to lift a 285-kg piano to a sixth-story window 16.0-m above?

SOLUTION:

The amount of work required is the increase in potential energy: W = mg Dy.

We find the time from

P = W/t = mg Dy/t;

1750 W = (285 kg)(9.80 m/s2)(16.0 m)/t, which gives t =25.5 s.

8)      If a car generates 18 hp (1 horse power = 746 Watts) when traveling a steady 90 km/h, what must be the average force exerted on the car due to friction and air resistance?

SOLUTION:

We find the equivalent force exerted by the engine from

P = Fv;

(18 hp)(746 W/hp) = F(90 km/h)/(3.6 ks/h), which gives F = 5.4 ´ 102 N.

At constant speed, this force is balanced by the average retarding force, which must be5.4 ´ 102 N.

## B LEVEL PROBLEMS

1)      What is the minimum work needed to push a 1000-kg car 300 m up a 17.70 incline? (a) ignore friction. (b) Assume the effective coefficient of friction is 0.25.

SOLUTION:

The minimum work is needed when there is no acceleration.

(a)  From the force diagram, we write SF = ma:

y-component: FNmg cos q = 0;

x-component: Fminmg sin q = 0.

For a distance d along the incline, we have

Wmin  = Fmind cos 0° = mgd sin q (1)

= (1000 kg)(9.80 m/s2)(300 m) sin 17.5°

=8.8 ´ 105 J.

(b)  When there is friction, we have

x-component: Fminmg sin q – mkFN = 0,or

Fmin = mg sin qmkmg cos q= 0,

For a distance d along the incline, we have

Wmin    = Fmind cos 0° = mgd (sin qmk cos q)(1)

= (1000 kg)(9.80 m/s2)(300 m)(sin 17.5° + 0.25 cos 17.5°) =1.6 ´ 106 J.

2)      At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. It was a rainy day and the coefficient of friction was estimated to be 0.42.use the data to determine the speed of the car when the driver slammed on (and locked) the brakes.(Why does the car’s mass not matter?)

SOLUTION:

On a level road, the normal force is mg, so the kinetic friction force is mkmg . Because it is the (negative) work of the friction force that stops the car, we have

W = Dke;

– mkmg d = ½mv2 – ½mv02;

– (0.42)m(9.80 m/s2)(88 m) = – ½mv02, which gives v0 =27 m/s (97 km/h or 60 mi/h).

Because every term contains the mass, it cancels.

3)      A 55-kg hiker starts at an elevation of 1600 m and climbs to the top of a 3100-m peak. (a) What is the hiker’s change in potential energy? (b) What is the minimum work required of the hiker? (b) Can the actual work done be more than this? Explain why.

SOLUTION:

(a)  With the reference level at the ground, for the potential energy change we have

Dpe = mg Dy = (55 kg)(9.80 m/s2)(3100 m – 1600 m) = 8.1 ´ 105 J.

(b)  The minimum work would be equal to the change in potential energy:

Wmin =Dpe8.1 ´105 J.

(c)    Yes, the actual work will be more than this. There will be additional work required for any kinetic energy change, and to overcome retarding forces, such as air resistance and ground.

4)      A projectile is fired at an upward angle of 45.00 from the top of a 265-m cliff with a speed of 185 m/s. What will be its speed when it strikes the ground below? (Use conservation of energy).

SOLUTION:

We choose the potential energy to be zero at the ground (y = 0).Energy is conserved, so we have

ekeipekefpe;

½mvi2 + mgyi = ½mvf2 + mgy;

½m(185 m/s)2 + m(9.80 m/s2)(265 m) = ½mvf2 + m(9.80 m/s2)(0), which gives vf199 m/s.

5)      A small mass slides without friction along a looped apparatus as shown. If the object remains on the track, even at the top of the circle (whose radius is r), from what minimum height h must it be released?

SOLUTION:

With y = 0 at the bottom of the circle, we call the start point A, the bottom of the circle B, and the top of the circle C.

At the top of the circle we have the forces mg and FN, both downward, that provide the centripetal acceleration:

mg + FN = mvC2/r.

The minimum value of FN is zero, so the minimum speed at C is found from

vCmin2 = gr.

From energy conservation for the motion from A to C we have

keApeAkeCpe;

0 + mgh = ½mvC2 + mg(2r),

thus the minimum height is found from

gh = ½vCmin2 + 2gr = ½gr + 2gr, which gives h = 2.5r.

6)      A 90-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15-m the floor is frictionless, and for the next 15-m the coefficient of friction is 0.30.What is the final speed of the crate?

SOLUTION:

On the level the normal force is FN= mg, so the friction force is Ffr = µkmg.

For the work-energy principle, we have

WNC = Dke + Dpe = (½mvf2 – ½mvi2) + mg(hfhi);

F(L1 + L2) –µkmg L2 = (½mvf2 – 0) + mg(0 – 0);

(350 N)(15 m + 15 m) – (0.30)(90 kg)(9.80 m/s2)(15 m) = ½ (90 kg)vf2,

which gives vf =12 m/s.

7)      During a workout, the football players at State U ran up the stadium stairs in 61 s. The stairs are 140 m long and inclined at an angle of 300.If a typical player has a mass of 105 kg, estimate the average power output on the way up. Ignore any frictional forces.

SOLUTION:

The work done increases the potential energy of the player. We find the power from

P  = W/t = Dpe/t = mg(hfhi)/t

= (105 kg)(9.80 m/s2)[(140 m) sin 30° – 0]/(61 s) = 1.2 ´ 103 W(about 1.6 hp).

# A LEVEL PROBLEMS

1)      A 220-kg load is lifted 21.0-m vertically with an acceleration a = 0.150 g (where g = 9.80 m/s2) by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, (e) the final speed of the load assuming that it started from rest.

SOLUTION:

(a)  From the force diagram we write SFy = may:

Fmg = ma;

F– (220 kg)(9.80 m/s2) = (220 kg)(0.150 m/s2),

which gives FT =2.48 ´ 103 N.

(b)  The net work is done by the net force:

Wnet    = Fneth = (Fmg)h

= [2.48 ´103 N – (220 kg)(9.80 m/s2)](21.0 m) = 6.79 ´103 J.

(c)  The work is done by the cable is

Wcable   = FTh

= (2.48 ´103 N)(21.0 m) =5.21 ´104 J.

(d)  The work is done by gravity is

Wgrav    = – mgh

= – (220 kg)(9.80 m/s2)(21.0 m) =– 4.53 ´104 J.

Note that Wnet = Wcable + Wgrav.

(e)  The net work done on the load increases its kinetic energy:

Wnet = Dke = ½mv2 – ½mv02 ;

6.79 ´103 J = !(220 kg)v2 –0, which gives v =7.86 m/s.

2)      What should be the spring constant k of a spring designed to bring a 1200-kg car to rest from a speed of 100 km/h so that the occupants undergo a maximum acceleration of 5.0 g?

SOLUTION:

The maximum acceleration will occur at the maximum compression of the spring:

kxmax = Mamax= M(5.0g), which gives xmax = 5.0Mg/k.

For the motion from reaching the spring to the maximum compression of the spring,

we use energy conservation:

keipespringi kefpespringf;

½Mv2 + 0 = 0 + ½kxmax2.

When we use the previous result, we get

½Mv2 = ½k(5.0Mg/k)2, which gives

k = 25Mg2/v2 = 25(1200 kg)(9.80 m/s2)2/[(100 km/h)/(3.6 ks/h)]2 =3.7 ´103 N/m.

3)      A 0.520-kg wooden block is firmly attached to a spring (k = 180 N/m).It is noted that the block-spring system, when compressed 5.0-cm and release, stretches out 2.3-cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

SOLUTION:

On the level the normal force is FN= mg, so the friction force is Ffr = µkmg.

The block is at rest at the release point and where it momentarily stops before turning back.

For the work-energy principle, we have

WNC = Dke + Dpe = (½mvf2 – ½mvi2) + (½kxf2 – ½kxi2);

µkmg L = (0 – 0) + ½k(xf2xi2 );

µk(0.520 kg)(9.80 m/s2)(0.050 m + 0.023 m) = ½ (180 N/m)[(0.023 m)2 – (– 0.050 m)2],

which gives µk =0.48.

4)      A 1000-kg car has a maximum power output of 120 hp. How steep a hill can it climb at a constant speed of 70 km/h if the frictional forces add up to 600 N?

SOLUTION:

From the force diagram for the car, we have:

x-component:   FFfr= mg sin q.

Because the power output is P = Fv, we have

(P/v) – Ffr= mg sin q.

The maximum power determines the maximum angle:

(Pmax/v) – Ffr= mg sin qmax ;

(120 hp)(746 W/hp)/[(70 km/h)/(3/6 ks/h)] – 600 N =

(1000 kg)(9.80 m/s2) sin qmax

which gives sin qmax = 0.409,orqmax = 24°.

5)      A small particle drops onto a spring as shown in the diagram. It compresses the spring a distance of 0.25 m before coming to a stop. Calculate the spring constant of the spring.

SOLUTION:

The energy of the mass is purely potential before it was dropped.

Ei = mgyi = (0.10kg)(9.80 m/s2)(1.0 m) = 0.98 J

The energy is purely potential once the spring is fully compressed, however it has both gravitational potential and elastic potential energy.

Ef = mgyf + ½ ky2f = (0.10 kg)(9.80 m/s2)(-0.25 m) + (0.5)k(-0.25 m)2 = -0.245 J + .03125 k

Conservation of energy applies:

0.98 J = -0.245 J + .03125 k

k = 39.2 N/m