A 1,080-N crate is being pushed across a level floor at a constant speed by a fo

Question

A 1,080-N crate is being pushed across a level floor at a constant speed by a force of 210 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?
k =

(b) If the 210-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).
m/s2

In part a) I got 0.171 which is right but I can't figure out the second part.

I wrote.

(210cos20-0.171 (1008.1758)) / 110.2041

= 0.2263 m/s^2

Where did I go wrong?

Explanation / Answer

Part 2. Now the normal force (Fn) = 1080- 210sin20 = 1008.175 N
Frictional force = uFn = 0.171 x 1008.175 ( since part1 u = 0.171) = 172.398N
Net force F' = F-f = 210cos20 - 172.398 = 24.93 N Acceleration = F/m = 24.93/(1080/9.8) = 0.226 m/s2

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