Question

A 120N box is being pushed across the surface that has a frictional force of 3N. If the box is being pushed with a constant force of 13N, what is the net force on the box? Remember, draw your free body diagram.

What is the acceleration of the box in the question above?

Answer 1

Net Force = 10 N

acceleration:
`a=0.817 (m)/(s^2)`

Step-by-step explanation:

1) The net force on the box is the applied force (F = 13 N) minus the friction force (f = 3 N), Therefore, net force = 10 N

See attached free body diagram.

2) To find the box's acceleration, we need first to find the mass of the box (since they are giving us its weight in Newtons). To do such we use the equation for weight and solve for the box's mass:

`weight=m*g\\120 N = m*9.8(m)/(s^2) \\(120)/(9.8) kg = m\\m=12.24 kg`

Now we can find the box's acceleration by using the equation for the net force:

`F=m*a\\10N = 12.24 kg * a\\(10)/(12.24) (m)/(s^2) =a\\a=0.817 (m)/(s^2)`