Question

A 1200-N crate is being pushed across a level floor at a constant speed by a force of 260 N at an angle of 20.0° below the horizontal.
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 260 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in the figure, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
For (a) I got .489, which is wrong. Since I cannot figure out (a) I cannot figure out (b).

Answer 1

For (a), the kinetic friction coefficient Uk, you need to divide the motion-producing force component by the total force normal to the plane
Uk = 260cos20/(1200 + 260 sin20)
= 0.1896

Answer 2

whats the equation I should use for (b) because I'm getting the wrong answer.

Answer 3

(b) The motion-producing force component remains 260cos20, but the normal component applied to the floor becomes 1200 -260sin20.

Then compute the net force along the direction of motion:
260cos20 - 0.1896*(1200 -260sin20)

Divide that by the mass (1200/g)to get the accleration.

Answer 4

.138