A 1,080 N create is being pushed across a level floor at a constant speed by a f

Question

A 1,080 N create is being pushed across a level floor at a constant speed by a force f of 220 N at an angle of 20.0 degree below the horizontal, as shown in the figure (a) below What is the coefficient of kinetic friction between the crate and the floor? If the 220 H force is instead pulling the block at an angle of 20.0 degree above the horizontal. as shown in the figure what will be the acceleration of the crate? Assume that the coefficient of friction n the same as that found m part (a). Your response differs from the correct answer by more than 10 percentage Double check your calculation. m/s

Explanation / Answer

a) Crate is moving with constant velocity hence a = 0
And Fnet on crate = ma = 0

in vertical,

N - Fsin20 - mg = 0

N = 220sin20 + 1000 = 1075.24 N

friction force, f = us N

along the surface,

Fcos20 - f = 0

220cos20 = us 1075.24

us = 0.192 ........Ans

b) now F is above the horizontal,

in vertical, N - mg + Fsin20 = 0

N = 1000 - 220sin20 = 924.76 N

f = us N = 0.192 x 924.76 = 177.55 N

In horizontal:

Fnet = ma

Fcos20 - f = ma

220cos20 - 177.55 = (1000/9.8)a

a = 0.29 m/s^2

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