Question

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. What is the acceleration of the crate? (Hint: remember to find net force first)

a=
m/s2

Answer 1

6.0 m/s2

Answer 2

Net force = Applied force - Force of friction
Net force = 1860 N - 480 N
Net force = 1380 N to the right

Answer 3

Net force = mass x acceleration

We substitute the values:

1380 N = 230 kg x acceleration

Solving for the acceleration:

acceleration = 1380 N / 230 kg

acceleration = 6 m/s^2 (rounded to the nearest whole number)

Therefore, the acceleration of the crate is 6 m/s^2.