137k views
5 votes
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?

1 Answer

5 votes

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.

User Paul Polash
by
5.3k points