Calculating Work and Friction in a Box on a Flat Floor

[gcharles_42]

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

 

[Yosty22]

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

 

[gcharles_42]

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

 

[Yosty22]

I believe so, yes.

 

[gcharles_42]

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

 

[Yosty22]

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.

 

[gcharles_42]

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

 

[gcharles_42]

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?

 

[SteamKing]

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

 

[gcharles_42]

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?