Homework #5

Solutions

13.(a)From the force diagram, we write F = ma:

y-component: FN – mg cos  = 0;

x-component: – F – kFN + mg sin  = 0.

Thus we have

F= – kFN + mg sin  = – kmg cos  + mg sin 

= mg (sin  – k cos )

= (380 kg)(9.80 m/s2)(sin 27° – 0.40 cos 27°) = 3.6102 N.

(b)Because the piano is sliding down the incline, we have

WF = Fd cos 180° = (3.6102 J)(3.5 m)(– 1) = – 1.3103 J.

(c)For the friction force, we have

Wfr = kmg cos dcos 180°

= (0.40)(380 kg)(9.80 m/s2)(cos 27°)(3.5 m)(– 1) = – 4.6103 J.

(d)For the force of gravity, we have

Wgrav= mg d cos 63°

= (380 kg)(9.80 m/s2)(3.5 m)(cos 63°) = 5.9103 J.

(e)Because the normal force does no work, we have

Wnet= Wgrav + WF + Wfr + WN

= 5.9103 J– 1.3103 J– 4.6103 J 0 = 0.

18.The magnitudes of the vectors are

A = (Ax2 + Ay2 + Az2)1/2 = [(6.8)2 + (4.6)2 + (6.2)2]1/2 = 10.3.

B = (Bx2 + By2 + Bz2)1/2 = [(8.2)2 + (2.3)2 + (– 7.0)2]1/2 = 11.0.

We find the angle from the scalar product:

A·B = AxBx + AyBy + AzBz = AB cos ;

(6.8)(8.2) + (4.6)(2.3) + (6.2)(– 7.0) = (10.3)(11.0) cos , which gives cos  = 0.203,  = 78°.

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

W = ΔK;

– kmgd = ½ mv2 – ½ mv02;

– (0.38)m(9.80 m/s2)(78 m) = – ½ mv02, which gives v0 = 24 m/s (87 km/h or 54 mi/h).

Because every term contains the mass, it cancels.

14.We choose y = 0 at the level of the trampoline.

(a)We apply conservation of energy for the jump from the top of

the platform to the trampoline:

E = K1+ U1 = K2 + U2;

½ mv12 + mgH = ½ mv22 + 0;

½ m(5.0 m/s)2 + m(9.80 m/s2)(2.0 m) = ½ mv22,

which gives v2 = 8.0 m/s.

(b)We apply conservation of energy from the landing on the trampoline

to the maximum depression of the trampoline. If we ignore the small

change in gravitational potential energy, we have

E = K2+ U2 = K3+ U3;

½ mv22 + 0 = 0 + ½ kx2;

½ (75 kg)(8.0 m/s)2 = ½ (5.2104 N/m)x2,

which gives x = 0.30 m.

This will increase slightly if the gravitational potential energy is taken into account.

27.(a)We find the normal force from the force diagram for the ski:

y-component:FN1= mg cos ;

which gives the friction force: Ffr1 = kmg cos .

For the work-energy principle, we have

WNC= ΔK + ΔU = (½mvf2 – ½mvi2) + mg(hf – hi);

– kmg cos L = ( ½mvf2 – 0) + mg(0 – L sin );

– (0.090)(9.80 m/s2) cos 20° (100 m) =

½ vf2 – (9.80 m/s2)(100 m) sin 20°,

which gives vf = 22 m/s.

(b)On the level the normal force is FN2= mg, so the

friction force is Ffr2 = kmg.

For the work-energy principle, we have

WNC= ΔK + ΔU = (½mvf2 –½mvi2) + mg(hf – hi);

– kmgD = (0 – ½ mvi2) + mg(0 – 0);

– (0.090)(9.80 m/s2)D = – ½ (22.5 m/s)2 ,

which gives D = 2.9102 m.

66.For the work-energy principle applied to coasting down the hill a distance L, we have

WNC= ΔK + ΔU = ( ½ mvf2 – ½ mvi2) + mg(hf – hi);

– FfrL = ( ½ mv2 – ½ mv2) + mg(0 – L sin ), which gives Ffr = mg sin .

Because the climb is at the same speed, we assume the resisting force is the same.

For the work-energy principle applied to climbing the hill a distance L, we have

WNC= ΔK + ΔU = (½mvf2 – ½ mvi2) + mg(hf – hi);

FL – FfrL = ( ½ mv2 – ½ mv2) + mg(L sin  – 0);

(P/v) – mg sin  = mg sin , which gives

P = 2mgv sin  = 2(75 kg)(9.80 m/s2)(5.0 m/s) sin 7.0° = 9.0102 W (about 1.2 hp).

NOTE: I told a few people to ignore friction for this problem, that was my bad.. though the answer for that case would be 4.5 x 102W

7.(a)We choose downward as positive. For the fall we have

y = y0 + v0t1 + ½ at12;

h = 0 + 0 + ½ gt12, which gives t1 = (2h/g)1/2.

To reach the same height on the rebound, the upward motion must be a reversal of the downward

motion. Thus the time to rise will be the same, so the total time is

ttotal = 2t1 = 2(2h/g)1/2 = (8h/g)1/2.

(b)We find the speed from

v = v0 + at1 = 0 + g(2h/g)1/2 = (2gh)1/2.

(c)To reach the same height on the rebound, the upward speed at the floor must be the same as the

speed striking the floor. Thus the change in momentum is

Δp = m(– v) – mv = – 2m(2gh)1/2 = = – (8m2gh)1/2 (up).

(d)For the average force on the ball we have

F = Δp/Δt = – (8m2gh)1/2/(8h/g)1/2 = – mg (up).

Thus the average force on the floor is mg (down), a surprising result.

16.For the collision we use momentum conservation:

x-direction: m1v1 + 0 = m1v1 cos 1 + m2v2 cos 2;

m(17 m/s) = mv1 cos 45° + mv2 cos 30°;

y-direction: 0 + 0 = – m1v1 sin 1 + m2v2 sin 2;

0 = – mv1 sin 45° + mv2 sin 30°.

The mass cancels, so we solve these two equations for

the two unknowns:

v1 = 8.8 m/s; v2 = 12.4 m/s.

46.On the horizontal surface after the collision, the normal force on the joined cars is FN = (m + M)g.

We find the common speed of the joined cars immediately after the collision by using the

work-energy principle for the sliding motion:

Wfr = ΔK;

– µk(m + M)gd = 0 – ½ (M + m)V2;

(0.40)(9.80 m/s2)(4.8 m) = ½ V2, which gives V = 6.13 m/s.

For the collision, we use momentum conservation:

mv + 0 = (m + M)V;

(0.95103 kg)v = (0.95103 kg + 2.2103 kg)(6.13 m/s), which gives v = 20 m/s (73 km/h).

64.We know from the symmetry that the center of mass lies

on a line containing the center of the plate and the center

of the hole. We choose the center of the plate as origin

and x along the line joining the centers. Then yCM = 0.

A uniform circle has its center of mass at its center.

We can treat the system as two circles:

a circle of radius 2R, density  and

mass p(2R)2 with x1 = 0;

a circle of radius R, density –  and

mass – pR2 with x2 = 0.80R.

We find the center of mass from

xCM= (m1x1 + m2x2)/(m1 + m2)

= [4pR2 (0) – pR2 (0.80R)]/(4pR2 – pR2)

= – 0.27R.

The center of mass is along the line joining the centers 0.07R outside the hole.