A 1,100-N crate is being pushed across a level floor at a constant speed by a force of 290 N at an angle of 20.0° below the horizontal, as shown in the figure (a) below.
Image for A 1,100-N crate is being pushed across a level floor at a constant speed by a force of 290 N at an angle of 20
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 290-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).

The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing those surfaces from moving. Typically, it is represented by the Greek letter mu (μ). In terms of math, is equal to μ = F/N, where F stands for frictional force and N for normal force.

Given the weight of the crate = 1100

Hence Mass of the crate m = 1100/9.8 = 112.245 kg

a) Since it is moving with a constant velocity, the net force acting on the block is zero

Fnet = 290cos 20 - μ*(290sin20 + 1100)

---> μ = 290cos 20/(290sin20 + 1100) = 0.227

b) Net force acting on the block = 290cos 20 - 0.227*(1100 - 290sin20) = 45.08 N

Acceleration of the block = F/m = 45.08/112.245 = 0.4 m/s2

To learn more about the coefficient of friction the link is given below:

https://brainly.com/question/17580716?

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