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# 13.5kg crate at rest on 27degree incline.wat force of friction is required to keep crate at a constant velocity once it begins sliding down?

B) what it’s coefficient of friction if the crate is sliding down 5m/s?

### 3 Answers

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- Anonymous1 month ago
For an object sliding down an incline,

acceleration a = g(sinΘ - µ*cosΘ)

"constant velocity" means a = 0. So

0 = 9.8*(sin27º - µ*cos27º)

µ = sin27º / cos27º = tan27º = 0.510

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- oldschoolLv 72 months ago
We know the component of weight in line with the incline = mg*sin27 = 13.5*9.8*sin27 = 60N

Since there is no acceleration, Fnet = 0 meaning the weight in line with the incline is exactly countered by the friction.

m*g*sin27 = m*g*µ*cos27

sin27/cos27 = tan27 = µ = 0.51 <<<<<<<

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- Anonymous2 months ago
Force of friction must equal the force of gravity acting parallel to the slope = mass * g * sin 27 deg.

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