Answers for Homework Handout for unit 4

Work and Energy

1. Here we useW = ~F � ~D = FD cos θ:

a)W = (30 N)(12 m) cos 0◦ = 360 J.

b) W = (710 N)(3.2 m) cos 31◦ = 1.95× 103 J.

c) W = (55 N)(10 m) cos 180◦ = −5.5× 102 J.

2. One key phrase is “constant velocity,” meaning thatFnet = 0 and thusWnet = 0. (Thework-energy theorem gives the same result, because∆K = 0, also.) That answers part (b).For part (a), we find the applied forceFapp that is balanced by the kinetic frictional force,Ff = µkFN . From Newton’s second law, we determine thatFN = Fg = mg = 833 N. Sothe work done by the pusher is 1.76×103 J.

3. Here we assume that the patient starts at rest and finishes at rest, so thatWnet = 0 by thework-energy theorem. The only forces doing work are gravity(Fg) and the applied force(Fapp). The work done by gravity depends only on the change in height,Wg = −mg(∆y),soWapp must be equal and opposite since the net work is zero:Wapp = mg(∆y) = 330 J.

4. The spring constant comes fromFapp = k(∆x); k = (200 N)/(0.025 m) = 8000N/m Thenthe work doneW = 1

2k(∆x)2 = 7.1 J.

5. battery: chemical/electrical energy

sling shot: elastic (spring) potential energy

wood and oxygen: chemical

climber on mountain: gravitational potential energy

spinning top: (rotational) kinetic energy

pot of hot water: thermal (kinetic) energy

x-rays: electromagnetic energy (not really “storage” per se, however)

6. From the top to the bottom, the gymnast’s center of mass moves downward by 2 m (∆y =−2 m), allowing gravity to do positive work (i.e., turning gravitational potential energy intokinetic energy). We assume that the friction of the hands on the bar does negligible workand can be ignored. The tension force in the arms also does zero work because it is alwaysperpendicular to the direction of motion. By the work-energy theorem,

Wnet = Kf −Ki

the only force doing work is gravity, andK = 12mv2, so

Wg =1

2mv2

f −1

2mv2

i

and solve forvf after substitutingWg = −mg(∆y). Note that the mass divides out of theequation! (vf = 6.37 m/s)

7. In this problem we can use the work-energy theorem to find the answer to (b) first and usethat answer to find (a). In class you also saw how to do part (a) by using forces. The onlyacting forces are gravity and the normal force (FN ), andFN does zero work because it isperpendicular to the displacement (cos 90◦ = 0). Work-energy tells us thatWnet = Kf −Ki,and we know thatKf = 0 (comes to rest at the top of the incline). The only work comesfrom Wg = −mg(∆y), where∆y comes from the geometry of the incline:sin 10◦ = (∆y)

L

(L = length of incline = 0.80 m). So∆y = 0.138m and Ki = 1.01 J. Then for part (a),vi = 1.6 m/s (from Ki = 1

2mv2

i ).

8. Another use for the work-energy theorem! We know the mass of the block and its initialand final speeds, so we knowKi = 0 and Kf = 1

2mv2

f . The work done by friction isWf = FfD cos 180◦ (whereD = 0.9 m andFf = µFN = µFg = µmg by Newton’s lawandµ = 0.4). The spring work isWs = 1

2k(∆x)2, where∆x = 0.20 m.

Wnet = Kf −Ki

Wf + Ws =1

2mv2

f − 0

−µmgD +1

2k(∆x)2 =

1

2mv2

f

Plugging in numbers and solving for the spring constant, we getk = 8150 N/m.

Power and Efficiency

9. The power (1200 W) is the work done per time:P = W/(∆t). So we must find the workdone to calculate the time. The work done by the motor is the work against gravity, soWm = −Wg = mg(∆y) = 1.66 × 104 J, so∆t = 13.7 s. [The only forces doing work arethe applied force and gravity, and the net work is zero (constant velocity).]

10. Efficiency isε = Wout/Win. Here,Wout = 72 J andWin = 600 J, soε = 0.12 (or 12%).

11. Win = 48000 J andWout = Win −Wwaste = 15000 J. Soε = 0.31 (or 31%).

Momentum and Impulse

12. a) The momentum gained by the wall is the opposite of the change in momentum of theball (by conservation of momentum).∆~p = ~pf − ~pi. If we take the initial direction of theball to be the positivex direction, then~pi = m~vi = (0.5 kg)(12 m/s x) = 6 kg m/s x, and~pf = −6 kg m/s x. So∆~pball = −12 kg m/s x and∆~pwall = −∆~pball = 12 kg m/s x.

b) The impulse is~Fave(∆t) = ∆~p. So for∆t = 0.01 s, ~Fave = 1200 N.

13. Momentum is conserved because the net external force is zero, so that~ptot,f = ~ptot,i. Theastronaut and wrench are initially “at rest” (with respect to the spacecraft), so~ptot,i = 0(va,i = 0 andvw,i = 0). So we have for the final total momentum (using~vw,f = 5 m/s x):

~ptot,f = 0

ma~va,f + mw~vw,f = 0

ma~va,f = −mw~vw,f

~va,f =−mw~vw,f

ma

= −0.083 m/s

(The negative sign indicates the opposite direction from the wrench.)

14. Again,~ptot,f = ~ptot,i by momentum conservation (zero net external force on the system ofthe student [st] and snowball [sb]). In this case~vst,i = 0 and~vsb,i = 22 m/s x. For the finalmomentum, the two masses stick together and move with a common velocity~vf :

~ptot,f = ~ptot,i

mtot~vf = mst~vst,i + msb~vsb,i

(mst + msb)~vf = 0 + msb~vsb,i

Then~vf = 0.13 m/s.

15. a) Same direction:~ptot,f = ~ptot,i

mtot~vf = m1~v1 + m2~v2

(60, 000 kg)~vf = (40, 000 kg)(3.0 m/s x) + (20, 000 kg)(5.0 m/s x)

Then~vf = 3.7 m/s.

b) Opposite direction:~ptot,f = ~ptot,i

mtot~vf = m1~v1 + m2~v2

(60, 000 kg)~vf = (40, 000 kg)(3.0 m/s x) + (20, 000 kg)(−5.0 m/s x)

Then~vf = 0.33 m/s.