Question

A large box has a mass of 500kg and the coefficient of static friction for the box and the floor is 0.45, and the coefficient of kinetic friction is 0.30.
a. What is the minimum horizontal force needed to get the box moving?
b. If you continue to push with that force, what will the acceleration of the box be?
Please help asap. Thank you.

Answer 1

A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.

b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight). 

F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N

a = (735.75 N)/(500kg)= 1.4715 m/s^2