A crate of mass 69.8 kg rests on a level surface, with a coefficient of kinetic

Question

A crate of mass 69.8 kg rests on a level surface, with a coefficient of kinetic friction 1.14. You push on the crate with an applied force of 1.046. What is the magnitude of the crates acceleration as it slides?
B)Take the same crate of mass 69.8 kg, and the same coefficient of kinetic friction 1.14, but now place the crate on an inclined surface, slanted at some angle above the horizontal. Now instead of pushing on the crate, you let it slide down due to gravity. What must angle be, in order for the crate to slide with the same acceleration it had in part?
C) Same situation as in part b, but now with no friction. Once again, what must the angle be, in order for the crate to slide with the same acceleration it had in part?
D) Different situation now. You're out in space, on a rotating wheel-shaped space station of radius 797 m. You feel planted firmly on the "floor", due to artificial gravity. The gravity you experience is Earth-normal, that is g=9.8. How fast is the space station rotating in order to produce this much artificial gravity?
E) What would the radius of the space station have to be, rotating at the rate you calculated in part (d), to produce artificial gravity of 30 m/s^2.?

Explanation / Answer

(A) N = m g = 69.8 x 9.8 = 684.04 N

f = uk N = 1.14 x 684.04 = 779.8 N

Fnet = m a

F - f = m a

1046 - 779.8 = 69.8a

a = 3.81 m/s^2

(B) N = m g cos(theta)

f = uk N = 1.14 m g cos(theta)

Fnet = f - m g sin(theta) = 0

1.14 m g cos(theta) = m g sin(theta)

tan(theta) = 1.14

theta = 48.7 deg

(C) m g sin(theta) = m a

a = g sin(theta)

3.81 = 9.8 sin(theta)

theta = 22.9 deg

(D) a = w^2 r

9.8 = w^2 x 797

w = 0.11 rad/s

(E) 30 = 0.11^2 r

r = 2439.8 m

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