A 900-N crate is being pushed across a level floor at a constant speed by a forc

Question

A 900-N crate is being pushed across a level floor at a constant speed by a force of 250 N at an angle of 20.0° below the horizontal. <BR><BR>(a) What is the coefficient of kinetic friction between the crate and the floor?_________________.<BR><BR>(b) If the 250 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)._____________________<BR>

Explanation / Answer

angle ? = 20 degrees Let force in horizontal direction = frictional force.
Fcos20 = µ(mg+Fsin20), where µ = coefficient of kinetic friction µ = Fcos20/(mg+Fsin20) µ = (250cos20) / (900 + 250sin20) = .238
Part 2. Now the normal force (Fn)= 900 - 250sin20 = 814.5 N
Frictional force = uFn = .238 x 814.5 = 194.2 N
Net force F' = F-f = 250cos20 - 194.2 = 40.8 N Acceleration = F/m = 40.8/(900/9.8) = .444 m/s2

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