# Work with velocity and force

• gcharles_42
In summary, a 49 kg box is pushed horizontally by a woman with a force of 628 N until it reaches a speed of 3 m/s. The coefficient of kinetic friction between the box and the floor is 0.60. The work done by the woman on the box is 220.5 J, while the CM-work done by the friction force on the box is 101.1631 J.

## Homework Statement

A box of mass 49 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0.60. A woman pushes horizontally against the box with a force of 628 N until the box attains a speed of 3 m/s.

What is the work done by the woman on the box?

and

What is the CM-work done by the friction force on the box?

## Homework Equations

W=F (delata x)
&
W(f)= uN (delta x)

## The Attempt at a Solution

Since work is in joules, I tried calculating it by multiplying force by velocity squared but that gave me a wrong answer. If I had the right answer for work I'd solve for delta x and use that to solve for work of friction

Try to think Kinematics: You know the mass, you know that it starts from rest, and you know the box's final speed.

Also CM-work is center of mass right? not centimeter?

I believe so, yes.

vf^2 = vo^2 + 2(a) d? seems to be what I'm looking for maybe. Do I use F=ma to splve for a? so 3^2 = 2 (d) 628/49. making d = 441/1256? so W= 628 (441/1256) = 220.5

I believe you can use f=ma to solve for acceleration. From there, use a kinematics equation to solve for distance in the x direction. Then you can multiply that by the force to solve for work.

But that's the KE, they're not the same are they?

So delta KE is 220.5, work by woman 220.5, and work by the force of friction = (u)mg (d) = .6(9.8)49( 441/1256) = 101.1631... ? Is that right?

gcharles_42 said:
But that's the KE, they're not the same are they?

That's why dimensional analysis is so handy.

Energy and work have the same derived units: W = (M*L^2/T^2)

If you apply dimensional analysis to your original supposition that W = F*V^2,
you would see that F = M*L/T^2 and V^2 = L^2/T^2, so W = M*L^3/T^4,
which isn't even close to the correct W = M*L^2/T^2

Yeah, I didn't use that supposition. I used vf^2 = vo^2 + 2(a) d instead to find distance... I just want to know if my answers for work of the woman and of the friction force are correct?