A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force Fvec of magnitude 6.1 N and a vertical force Pvec are then applied to the block. The coefficients of friction for the block and surface are ?s = 0.36 and ?k = 0.24. Determine the magnitude of the frictional force acting on the block if the magnitude of Pvec is 8.0 N. Determine the magnitude of the frictional force acting on the block if the magnitude of Pvec is 10.0 N. Determine the magnitude of the frictional force acting on the block if the magnitude of Pvec is 12.0 N.

Explanation / Answer

a) W= mg= 24.5 N, P =8 N

N = mg- p = 16.5 N

Fr = us * N = 0.36 *16.5 = 5.94 N

Horizontal force = 6.1 so block moves

Friction = uk * 0.24 * 16.5 = 3.96 N

b) Fr = us(mg-P) = 5.22 N

So kinetic friction acts

Fr = uk * 14.5 = 3.48 N

c) Fr = us*N = 0.36 * (24.5-12) =4.5

Horizontal force = 6.1 N

so kinetic friction acts

Fr = uk*N = 0.24 *12.5 = 3 N

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