A 1.5 kg block is pulled across a horizontal surface that has a coefficient of kinetic friction of 0.60. a. What is the force of friction exerted by the surface on the block?

Assuming the block is pulled with constant speed, there is no net force acting on it, so that

n + (-w) = 0

p + (-f ) = 0

where n and w denote the magnitudes of the normal force and the block's weight (both acting in the vertical direction), and p and f denote the magnitudes of the pulling and friction forces (both in the horizontal direction).

f is proportional to n by a factor µ = 0.60, so that

f = 0.60 n

The block has weight

w = (1.5 kg) (9.80 m/s²) = 14.7 N

and hence n = 14.7 N.

Then the friction force has magnitude

f = 0.60 (14.7 N) ≈ 8.8 N

and opposes the direction of movement.

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