Physics 101 - Homework # 5 Solutions

• Problem 6-20
• Problem 6-62
• Problem 6-72
• Problem 7-3
• Problem 7-8
• Problem 7-23
• Problem 7-27
• Problem 7-32
• Problem 7-35
• Problem 7-59
• Problem 8-2
• Problem 8-5
• Problem 8-11

Chapter 6

Problem 20)

For a constant force the work is

W_F = F·r = [(2.7 N)i - (6.8 N)j)]·[(3.4 m)i + (1.2 m)j]

= (2.7)(3.4) J - (6.8)(1.2) J = 1.02 J

Problem 62)

Recall that power is the rate that work is being done. First convert the speed units:

(100 km/h)(1000m/km)/(3.6x10³s/h) = 27.8 m/s

The required work is

W = K = ½mv² - ½m(0)²

From

P = W/t

we find

t = W/P = ½(1200 kg)(27.8 m/s)²/[(80 hp)(746 W/hp)] = 7.8 s

Aside: this assumes that all of the power goes into accelerating the car, i.e. none is wasted...

Problem 72)

A joule is not a very big unit.

Chapter 7

Problem 3)

Notice that we are not told anything about the direction the ball is thrown! We would seem to need this if we were to use our projectile motion techniques to solve this. Instead, use energy conservation.

Convert the speeds: (95 mi/h)(1.61 km/mi)(1000 m/km)/(3600 s/h) = 42.5 m/s; similarly 120 mi/h = 53.7 m/s.

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

E = K_i + U_i = K_f + U_f

= ½mv_i² + mgy_i = ½mv_f² + mgy_f

= ½m(42.5 m/s)² + m(9.8 m/s²)(80 m) = ½m(53.7 m/s)² + m(9.8 m/s²)y_f

which gives (the mass m cancels out):

y_f = 25 m

Problem 8)

The potential energy for a Hooke's-law spring is U = ½kx², if we choose U=0 when x=0.

Then

1.565 J = ½(42.38 N/m)x²,

which gives x = 0.2718 m.

Because the energy is constant, the speed is maximum when U=0 :

E = K₁ + U₁ = K₂ + U₂

= 0 + 1.565 J = ½mv_max² + 0 = ½m(2.402 m/s)²

which gives m = 0.5425 kg.

Problem 23)

The potential energy is U(x) = x⁴ + ßx².

The equilibrium positions are local maxima or minima, and so have   dU/dx = 0   or

dU/dx = 4x³ + 2ßx = 0

from which we get

x=0    and

x = ±(-ß/2)^1/2 = ±[-(-3.0 J/m²)/2(26 J/m⁴)]^1/2 = ± 0.24 m

Problem 27)

Let's choose a coordinate system with y=0 at the surface of the sea. We are told we can neglect air resistance, and thus the mechanical energy is conserved:

E = (1/2)mv_i² + mgy_i = (1/2)mv_f² + mgy_f

Divide through by the common factor of m (which is good, since it wasn't give to us!), and then plugging in numbers we have

(1/2)(125 m/s)² + (9.8 m/s²)(68 m) = (1/2)v_f² + 0

which can be solved for v_f

v_f = 130 m/s

If the cannon is fired at an inclined angle, the change in gravitational potential energy is the same, so the kinetic energy change will be the same, and the final speed will be the same. The direction of the velocity, however, will be different.

Problem 32)

Let's call the start point A, the top of the intermediate peak point B, and the bottom of the hill C, with y=0 at the bottom of the hill. From energy conservation we have:

K_A+U_A = K_B+U_B = K_C+U_C

0 + mgh_A = ½mv_B² + mgh_B = ½mv_C² + 0

Thus

v_B² = 2g(h_A-h_B) =2(9.8 m/s²)(95 m - 32 m)

which gives

v_B = ±35.1 m/s

Also

v_C² = 2gh_A = 2(9.8 m/s²)(95 m)

which gives

v_C = ±43.2 m/s    (= 96 mi/h)

The neglect of resistive forces is not reasonable, no matter how well-waxed the skis; there is lots of air resistance at these speeds!

Problem 35)

Let's choose y=0 at the bottom.

a) For the motion from the initial point to point a we have, from energy conservation,

K_i + U_i = K_a + U_a

0 + mgh_i = ½mv_a² + 0

m(9.8 m/s²)(0.15 m) = ½mv_a²

which gives

v_a = 1.71 m/s

b) For the motion from point a to point b, taking into consideration the effect of friction, we have

W_f = K + U

since the change in total energy is the work done by the non-conservative forces (here friction). Thus

-µ_kmgx = ½mv_b² - ½mv_a² + 0

-(0.21)m(9.8 m/s²)(0.30 m) = ½mv_b² - ½m(1.7 m/s)²

   which gives

v_b = 1.29 m/s

c) For the motion from the initial point to the stopping point, we let D represent the total horizontal distance over which the friction force acts. Because the frictional force always opposes the motion we have

W_f = K + U - µ_kmgD

= (0 - 0) + (0 - mgh_i)

which gives

D = h_i/µ_k = (0.15 m)/(0.21) = 0.714 m = 71.4 cm

This corresponds to 71.4 cm - 2(30 cm) = 11.4 cm from point a.

Problem 59)

We choose the coordinate system shown (see diagram), with y=0 being the release point. Since the tension does no work (it is always perpendicular to the displacement), only conservative forces (gravity) act, and we can apply conservation of mechanical energy.

a) From the release point to the lowest location we have

K_i + U_i = K_f + U_f

0 + 0 = (1/2)mv² + mg(-L)

v² = 2gL = 2(9.8 m/s²)(1.0 m) = 4.4 m/s

b) From the release point to an angle with the vertical, we have

K_i + U_i = K_f + U_f

0 + 0 = (1/2)mv² + mg(-Lcos)

v² = 2gLcos = 2(9.8 m/s²)(1.0 m)cos45° = 3.7 m/s

c) The centripetal acceleration is provided by the tension and the component of the weight. At an angle with the vertical, we have

T - mgcos = mv²/L

Using the result from part b) for v², we have

T - mgcos = 2mgcos

or

T = 3mgcos

At the bottom of the swing, =0, so we have

T = 3(0.20 kg)(9.8 m/s²)cos(0°) = 5.9 N

When = 45° we have

T = 3(0.20 kg)(9.8 m/s²)cos(45°) = 4.2 N

The tension in the string is largest at the bottom of the swing.

Chapter 8

Problem 2)

Because the motion has constant acceleration, with positive=down, we find the velocity from

v = v₀ + at = 0 + (9.8 m/s²)(2.0 s) = 19.6 m/s

The momentum is p = mv = (0.800 kg)(19.6 m/s) = 15.7 kg· m/s   downward.

Note that the time of fall is

T_fall = [2y/g]^1/2 = [2(36.5 m)/(9.8 m/s²]^1/2 = 2.7 s,

which is greater than 2.0 s, so it has not hit the ground yet.

Problem 5)

a) During the firing we use momentum conservation:

p_ri + p_bi = p_rf + p_bf

0 + 0 = (7 kg)v_rf + (10 x 10^-3 kg)(700 m/s)

from which we get    v_rf = - 1.0 m/s

b) The energy transmitted to the shoulder comes from the decrease in kinetic energy of the rifle:

E = -(0 - ½mv²) = ½(7 kg)(-1.0 m/s)²

= 3.5 J

Problem 11)

For this one-dimensional collision we have J =p_f - p_i.

For the ball we get:

J_ball = (0.145 kg)(-45 m/s - 36 m/s)

= -11.7 kg·m/s ,    opposite to the original motion.

The average force is

F_av = J_ball/t

= (-11.7 kg·m/s)/(7.0 x 10^-4 s)

= -1.68 x 10⁴ N,   (opposite to the original motion).