Question

1.) Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N. The velocity of the crate is in the direction of the force Steve is applying to the crate. How much work is done by Steve?

2.) Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N. The velocity of the crate is in the direction of the force Steve is applying to the crate. How much work is done by friction?

3.) Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N. The velocity of the crate is in the direction of the force Steve is applying to the crate. How much work is done by gravity?

4.) Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N. The velocity of the crate is in the direction of the force Steve is applying to the crate. How much work is done by the normal force?

5.) Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate?

Answer 1

1)

W = F*d

= 200 N * 20 m

= 4000 J

Answer: 4000 J

2)

since object moves with constant velocity,

frictional force must be equal in magnitude to F

f = -F = -200 N

W = f*d

= -200 N * 20 m

= -4000 J

Answer: -4000 J

3)

force due to gravity is upward and displacement is in horizontal direction.

So, force and displacement are perpendicular.

So, work done = 0

Answer: 0

4)

Normal force is downward and displacement is in horizontal direction.

So, force and displacement are perpendicular.

So, work done = 0

Answer: 0

5)

net work done = sum of all work done

= 4000 - 4000 + 0 + 0

= 0 J

Answer: 0