You are pulling a crate across the floor.

Question

The crate has a mass of 77.0 kg. You exert a horizontal force of 121 N to the right. Friction exerts a force of 64.0 N to the left. What is the magnitude (fill in box) and direction of the acceleration of the crate?

Answer 1

This is a classic sum of forces problem. Whenever you have any number of forces all moving along the same axis you can add them up to get a net force. The equation is [total_force = force1 + force2 +...=(mass)*(acceleration)]. There can be any number of forces added together. In this case we are working with a horizontal axis.

There are 2 forces. The 121 N pushing force to the right and the friction directed to the left. Since they are both horizontal and opposite directions one should be negative. Convention would say friction should be negative.

Using the equation we get: total_force = 121N - 64.0N = (77 kg)*(acceleration)

Simplifying we get: (57 N)/(77 kg) = acceleration

The direction of acceleration is the same as the net force.

I hope this helped!

Answer 2

F = 121 N

f = magnitude of opposing force of friction = 64.0 N

m = mass of crate = 77.0 kg

a = (F - f) / m = (121 - 64.0)N / (77.0 kg) = ? m/s² (to the right)