A 10.0 kg crate initially at rest sitting on a floor with a coefficient of static friction of 0.20 is pushed with a force of 15 N for 4.0 s. What is the work done?

Question

1. 15 J
2. 60 J
3. 10 J
4. 600 J
5. 0

Answer 1

Hello Bella ,

The force of friction is 0.2*N

Where N= normal reaction = Mg= 10*g = 100N (for g= 10N/m2)

So force of friction = 20 N

which i more than force applied 15N

Thus the body will not move so the work done = 0 J

Option 5 is correct