Question

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the coefficient of kinetic friction is 0.200. What is the friction force on the box if a horizontal 140-N push is applied to it?

Answer 1

Answer:

147 N

Explanation:

static friction = frictional force max / normal

normal force = mg = 50 kg × 9.8 m/s² = 490 N

0.3 = Frictional force / 490 N

0.3 × 490 N = 147 N

The body will not move because the frictional force is greater than the force applied.

Answer 2

Answer:

The frictional force is 140 N

Explanation:

The static frictional force can be calculated with the expression below;

F = μmg

where F is the static frictional force

μ is the coefficient of static friction = 0.300

m is the mass of the body = 50 kg

g is the acceleration due to gravity = 9.8 m/$s^{2}$

Substituting into the equation above;

F = 0.300 x 50 kg x 9.8 m/$s^{2}$

F =  147 N

With a static force of 147 N which is greater than the applied force 140 N, the box will not move.

The frictional force on the box if a horizontal force of 140 N will be the same as the horizontal force. Therefore the frictional force would be 140.