A 1,320-N crate is being pushed across a level floor at a constant speed by a fo
ID: 1338671 • Letter: A
Question
A 1,320-N crate is being pushed across a level floor at a constant speed by a force vector F of 380 N at an angle of 20.0° below the horizontal, as shown in the figure (a) below.
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 380-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
Explanation / Answer
Normal force N = W+ Fsin theta
N = 1320 + 380 sin 20
N = 1450 N
frictional force Ff = F cos 20 = 380 cos 20 = 357.08 N
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A. Coeffcient of friction u = f/R = 357.08/1450 = 0.246
B. Accleration a = 380 cos 20 - u (W-380 sin 20)/m
a = (357.03 - 0.246 *(1320 -380 sin 20))/(1320/9.8)
a = 0.477 m/s^2
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