Question

A 70kg block is being pushed across a tabletop with a constant force of 350N exerted in the direction of travel. If the coefficient of kinetic fricton is 0.16, find the

A) Friction exerted on the block

B) Acceleration of the block

Answer 1

A)
The normal force on the block, N = M * g
Where M is the mass of the block and g is the acceleration due to gravity.
Frictional force, Ff = k * N
Where k is the coefficient of kinetic friction.

Ff = k * M * g
= 0.16 * 70 * 9.8
= 109.76 N
= 110 N

B)
Net force, Fn = Applied force - Frictional force
= 350 - 109.76
= 240.24 N

Acceleration, a = Fn/M
= 240.24 / 70
= 3.432 m/s2
= 3.4 m/s2