A 2.70 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.70 kg block is initially at rest on a horizontal surface. A horizontal force ModifyingAbove Upper F With right-arrow of magnitude 6.48 N and a vertical force ModifyingAbove Upper P With right-arrow are then applied to the block (see the figure). The coefficients of friction for the block and surface are ?s = 0.4 and ?k = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of ModifyingAbove Upper P With right-arrow is (a)9.00 N and (b)13.0 N. (The upward pull is insufficient to move the block vertically.)

Explanation / Answer

Part A)

The formula for friction is uFn

Fn = mg - P

Fn = (2.7)(9.8) - 9 = 17.46 N

If the friction was static, the force is .4(17.46) = 6.984 N

Since the applied force is less than this, the block can not move and the frictional force will be equal to the applied force.

That is 6.48 N

Part B)

The new normal force is

Fn = (2.7)(9.8) - 13 = 13.46 N

The static frictional force is .4(13.46) = 5.384 N

Since this is smaller than the applied force, the block will move, so the frictional force is based on the kinetic friction

F = (.25)(13.46)

F = 3.37 N

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