A) a crate with a mass of 33.1 kg rests on a level surface, with a coefficient o

Question

A) a crate with a mass of 33.1 kg rests on a level surface, with a coefficient of kinetic friction 1.74. you push on the crate with an applid forc of 0.827. what is the magnitude of the crates acceleration?

B)take the same crate with the mass of 33.1 kg,and the same coefficient of kinetic friction 1.74, but now place it 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 mus be the angle be in order for the rate to slide with the same acceleration it had in part (A) ?

C) same situation as in part (B), but now with no friction. what is the angle in order for the crate to slide with the same acceleration it had in part (A)?

D) different situation. your on a rotating wheel shaped space station with a radius of 672m. You are planted on the floor due to artificial gravity (g= 9.8m/s2). 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 30m/s2?

need answer for part D and E specifically

Explanation / Answer

m = 33.1 kg
mu = 1.74
a) friction,f = mu*mg = 1.74*33.1*9.8 = 564.4212 N
F = 827 N
Net F = 262.5788 N = 33.1*a [ newton's second law]
a = 7.9328 m/s/s
b) friction = mu*mg*cos(theta) = 564.4212cos(theta)
Downward force = mgsin(theta) = 324.38sin(theta)
Net force = 324.38sin(theta) - 564.4212cos(theta) = 33.1*7.9328 = 262.5756
105222.3844 - 105222.3844cos^2(theta) = 318571.291cos^2(theta) + 68945.945 + 296399.52cos(theta)
423793.67cos^2(theta) + 296399.52cos(theta) - 36276.44 = 0
cos(theta) = 0.1062493, -0.805645
theta = 83.9 deg or, 143.67
so theta = 83.9 deg
c) gsin(theta) = 7.9328
theta = arcsin(7.9328/9.8) = 53.963 deg
d) w^2 * r = 9.8
r = 672 m
w = 0.12076 rad/s

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