A 2.50 kg block is initially at rest on a horizontal surface. A horizontal force

Question

A 2.50 kg block is initially at rest on a horizontal surface. A horizontal force ModifyingAbove Upper F With right-arrow of magnitude 6.90 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)6.00 N and (b)10.0 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 = (2.5)(9.8) - 6 = 18.5 N

If the friction was static, the force is .4(18.5) = 7.4 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 6.9 N

Part B)

The new normal force is

Fn = (2.5)(9.8) - 10 = 14.5 N

The static frictional force is .4(14.5) = 5.8 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)(14.5)

F = 3.63 N

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