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1. Walker2 8.P.001. [236136] Calculate the work done by gravity as a 2.6 kg object is moved from point A to point B in Figure 8-15 along paths 1, 2, and 3.

Figure 8-15

path 1

Force of gravity is always pointing down

Work for each segment = F d cos(theta) = F_x d_x + F_y d_y

F_x = 0, F_y = -Mg

Work = (-Mg) 4m + 0 +(-Mg)(-1.0m) + 0 + (-Mg)(-1.0 m)

Work = (-Mg) (4.0 m –1.0 m - 1.0 m) = (- 2.6 kg)(9.8 m/s²)(2.0 m) = -51 kg m²/s²

[-51] J

path 2

Work = 0 +(-Mg)(2.0m) + 0 = (- 2.6 kg)(9.8 m/s²)(2.0 m) = -51 kg m²/s²

[-51] J
path 3

Work = (-Mg)(-1.0m) + 0 + (-Mg)(3.0m) = (-Mg)(2.0 m) =-51 kg m²/s²

[-51] J

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2. Walker2 8.P.002. [236137] Calculate the work done by friction as a 2.3 kg box is slid along a floor from point A to point B in Figure 8-16 along paths 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.25.

Force of friction is always opposing motion (theta=180 deg).

Path 1:

Work = F d cos(180) = m Mg d (-1)

Work = -m Mg ( 4.0 m + 4.0 m + 1.0 m + 1.0 m + 1.0 m)

Work = - (0.25) (2.3 kg) (9.8 m/s2) ( 11.0m)
[-62] J (along path 1)
[-28.2] J (along path 2)
[-39.5] J (along path 3)

Figure 8-16

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3. Walker2 8.P.007. [236141] Find the gravitational potential energy of an 79.5 kg person standing atop Mt. Everest, at an altitude of 8848 m. Use sea level as the location for y = 0.

U = mgh = (79.5 kg) (9.81 m/s2) (8848 m) = 6,900,000 kg m²/s²
[6.9] MJ

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4. Walker2 8.P.012. [236143] A 0.36 kg pendulum bob is attached to a string 1.2 m long. What is the change in the bob's gravitational potential energy as it swings from point A to point B in Figure 8-19, where = 40°?

Use the PIVOT point as the zero of potential energy.

At the bottom, U(B) = -mgL

At angle q: U(A) = -mgL cos(q)

Change in U = Final – Initial = U(B) – U(A) = -mgL – [– mgL cos(q) ]

Change in U = –mgL [ 1- cos(q)] = – (0.36 kg)(9.81 m/s2) (1.2 m) [ 1- cos(40)]

DU = –4.238 kg m²/s² [1-0.766 ] = -0.991 J
[-0.991] J

Figure 8-19

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5. Walker2 8.P.013. [236144] At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and falls through a vertical height of 2.53 m, what is her speed at the bottom of the slide?

E = K + U = constant

E(top) = 0 + mgh

E(bottom) = (1/2)mv² + 0

½ m v² = mgh

v²=2gh = 49.64 m²/s²
v = 7.05 m/s

[7.05] m/s

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6. Walker2 8.P.014. [236145] To enter the main pool at an amusement part, a swimmer uses a water slide which has a vertical height of 2.31 m. Find her speed at the bottom of the slide if she starts with an initial speed of 0.790 m/s.

E = K + U = constant

E(top) = (1/2)m v(initial)² + mgh

E(bottom) = (1/2)mv(final)² + 0

½ m v(final)² = mgh+ (1/2)m v(initial)²

v_f² = 2gh + v_i² = 2(9.81 m/ s²)(2.31m) + (0.790 m/s)²

v_f² = 45.32 m²/s² + 0.624 m²/s² = 45.95 m²/s² =
v = 6.78 m/s

[6.78] m/s

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7. Walker2 8.P.018. [236147] A 3.3 kg block slides with a speed of 1.1 m/s on a frictionless, horizontal surface until it encounters a spring.

(a) If the block compresses the spring 4.6 cm before coming to rest, what is the force constant of the spring?
E = K + U = constant
E(initial) = (1/2) m v² + 0
E(final) = 0 + (1/2) k x²
k = m v² /x ² =  (3.3 kg) (1.1 m/s)² / (0.046 m)² = 1887 kg /s ²= 1890 N/m
[1890] N/m
(b) What initial speed should the block have if it is to compress the spring by 1.4 cm?
[0.335] m/s

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8. Walker2 8.P.019. [236148] A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 37 m high. When it hits the ground at the base of the cliff the rock has a speed of 29 m/s.

(a) Assuming that air resistance can be ignored, find the initial speed of the rock.

E = K+U = (1/2) m v² + mgh

E(top of cliff) = E(bottom)

(1/2) m v_i² + mgh(cliff) = (1/2) m v_f² + 0

v_i² = v_f² - 2gh(cliff) = (29 m/s)² – 2(9.81 m/s²) (37 m) = 115.06 m²/s²

v_i = 10.7 m/s

[10.7] m/s
(b) Find the greatest height of the rock as measured from the base of the cliff.

At top of vertical motion, K=0 (rock was thrown straight up).

Mgh(top of motion) + 0 = E = (1/2) m v_i² + mgh(cliff)

h(top) = h(cliff) + (1/2) v_i² /g = 39 m + (0.5) (29 m/s)²/(9.81 m/s²) = 42.86 m

[42.9] m

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9. Walker2 8.P.040. [236161] An object moves along the x axis, subject to the potential energy shown in Figure 8-21. The object has a mass of 1.2 kg, and starts at rest at point A.

Figure 8-21

(a) What are the object's speeds at point B, at point C, and at point D?

E = Constant = K+U

K(A)+U(A) = K(B)+U(B)=K(C)+U(C)+…

K(A) = 0; U(A) = 10.0 J; E = 10.0 J

K(B) = E – U(B) = 10.0 J – 2.0 J = 8.0 J

(1/2) mv_B² = 8.0 J

v_B² = 2(8.0 J)/ (1.2 kg) = 13.33 m²/s²

v(B) = 3.65 m/s
[3.65] m/s (B)
[2.58] m/s (C)
[2.89] m/s (D)
(b) Which points are turning points for this object? (Select all that apply.)

These are the points where K=0, since U=E

[x] point A

[_] point B

[_] point C

[_] point D

[x] point E

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10. Walker2 8.P.041. [236162] A 1.2 kg object moves along the x axis, subject to the potential energy shown in Figure 8-21. If the object's speed at point B is 1.29 m/s, what are the approximate locations of its turning points?

Turning points occur when K=0 (U=E)

E = constant

E(B) = K(B) + U(B) = (1/2) (1.2 kg) (1.29 m/s)2 + 2.0 J = 3.0 J

Find the points on the graph where U=3.0 J
[1.6] m (lefthand turning point)
[2.3] m (righthand turning point)

Figure 8-21

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11. Walker2 8.P.043. [236163] A 23 kg child swings back and forth on a swing suspended by 2.3 m long ropes. Find the turning-point angle if the child has a speed of 0.81 m/s when the ropes are vertical (where = 0).

Turning points occur when K=0 (U=E)

E = constant

E(Bottom) = K(Bottom) + U(Bottom)

U(Bottom) = - mgR = -(23 kg) (9.81 m/s²) (2.3 m) = -518.9 J

K(Bottom) = (1/2) m v2 = 7.545 J

E = K+U = 7.545 J – 518.9 J = -511.4 J

U(q) = - mgR cos(q)

Find the angle such that K=0, U(q) = E

- mgR cos(q) = -511.4 J

cos(q) = 511.4 J / (23 kg * 9.81 m/s2 * 2.3 m) = 0.9854

q = 9.78 deg

Notice that the answer does not depend on the mass of the child

Cos(q) = [ (1/2 m v² – mgR] / [ -mgR] = 1 – v²/(2gR) = 0.9854

+/- [9.78]°

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12. Walker2 8.P.050. [236165] A 66 kg skier encounters a dip in the snow's surface that has a circular cross section with a radius of curvature of 12 m. If the skier's speed at point A in Figure 8-22 is 8.7 m/s, what is the normal force exerted by the snow on the skier at point B? Ignore frictional forces.
E = K + U
E(A) = K(A) + 0
K(A) = (1/2) (66 kg) (8.7 m/s)² = 2498 J
E(B) = K(B) + m g y(B)
y(B) = - 1.75 m
 K(B) = E(B) - m g y(B)
K(B) = K(A) - m g y(B)
divide both sides by m/2
v(B)² = v(A)²- 2 g y(B) = (8.7 m/s) ² - 2 (9.81 m/s²)(-1.75 m) = 110.0 m²/s ²
Acceleration at B = centripetal acceleration = v(B)² / r  = ( 110.0 m²/s²) / (12 m) = 9.169 m/s ²
Newton's 2nd law: F(Net) = m a
N- mg = ma
N = m (a + g) = (66kg)(9.169 m/s2 + 9.81 m/s2) = 1250 N

Once again N is NOT mg!!

[1.25] kN

Figure 8-22

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