Question

A crate is pushed 1.50m horizontally with a force of 2.40N under an opposing force of friction of 0.600N. What is the NET work done on the crate?

Answer 1

2.7 J

Step-by-step explanation:

Given that

Distance pushed through, s = 1.5 m

Force of the push, F(push) = 2.4 N

Opposing force of friction, F(f) = 0.6 N

See attachment for solution

The net work done is a sum of the calculated work done, and is given as

W(net) = 3.6 +(-0.9) + 0 + 0

W(net) = 2.7 J

Thus, the net work done is 2.7 J

Answer 2

2.7 J

Step-by-step explanation:

From the question,

The Net work done by the crate = Work done by the crate in moving if through a distance- work done against friction.

W = Fd-F'd.................... Equation 1

Where F = force applied to the crate, d = distance moved by the crate, F' = force of friction

Given: F = 2.4 N, F' = 0.6 N, d = 1.5 m

Substitute these values into equation 1

W = 2.4(1.5)-0.6(1.5)

W = 3.6-0.9

W = 2.7 J