Work (physics)

.kbf?

I got this homework problem yesterday. I can't figure it out for the life of me.

First off please please don't solve it for me. I just want a push in the right direction.

The problem goes something like this.

A 4.0 kg block is pushed up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 30 degrees with the horizontal. If the coefficient of friction between the block and the wall is 0.30 determine the work done by the force.

So far what I have is:

Force in the Y = Fsin(30) : Fy

Force in the X = Fcos(30) : Fx

Normal force between the wall and the block = -Fcos(30) : n

Force of Kinetic friction = 0.3 * (-Fcod(30)) : Fk

Fy - (Fk + mg) = ma (is this right?)

Fy - (Uk * n) - mg = ma

Fsin(30) = (ma - mg - (Uk * n))

F = (ma - mg - (Uk * n))/sin30

Work = f*Δy

W = [(ma - mg - (Uk * n))/sin30] * 4


Am I even thinking along the right lines?

(is a = 0 because it's a constant speed?)

physi_marc

You're in the right direction, yes.

Yes, a = 0 because the velocity is constant.

You can go further than you've gone, though, by eliminating the normal force. You should be able to figure out how to do this.

nuclearalchemist

Yeah, eliminate the normal force, and then just look at what happens in the other direction. You really only have one force acting on the block. Drawing a free force diagram would probably help in this case......

Sir Headless VII

Also you have messed up the negative signs between these two lines

Fy - (Uk * n) - mg = ma

Fsin(30) = (ma - mg - (Uk * n))

it should be Fsin(30) = (ma + mg + (Uk * n))

.kbf?

Sir Headless VII wrote: »

Also you have messed up the negative signs between these two lines

Fy - (Uk * n) - mg = ma

Fsin(30) = (ma - mg - (Uk * n))

it should be Fsin(30) = (ma + mg + (Uk * n))

Wow that's a stupid mistake. Thanks for catching that.

physi_marc wrote:

eliminating the normal force

I'm not exactly sure how to do that

I now have

W = 4[(mg+ (-Fcos30 * Uk))/sin30]

How do I get rid of the -Fcos30 is F is never specified? Solve for F and put that back into the equation?

edit: is it because f=ma and a=0?

Boutros

Write down the vertical equilibrium equation with everything written out in terms of F. You thought you solved for F, but n is still a function of F, so you haven't fully isolated F. And the work is the dot product of the force with the displacement, so you only care about the component of the force in the direction the box is moving so W=Fy*d=Fsin(30)d