How Is Work Calculated for a Force Applied at an Angle?

[PhysicsPete2564]

Homework Statement

In the figure below (image attached) a constant force of 82N is applied to a 3kg box at an angle of 53 degrees causing the box to move up a frictionless ramp. How much is work is done by that force (Fa) when the box has moved through a vertical distance of .15m?

Homework Equations

Work = Force X distance

The Attempt at a Solution

I thought this would be a simple problem. Work equals force X distance so I thought you would just multiply the force and distance vectors. Multiplying the vectors means using the dot product. So initially I just did:
W = (82N)*(.15m)*(Cos(53)). This equals 7.40 which is not the correct answer.

I found a problem similar to this on the web and apparently the way to do this problem is W = m*g*distance. This produces the right answer for the problem (4.41) but I don't understand the logic behind that equation. I thought since the force and distance are NOT equal in direction you need to multiply vectors. What am I missing here?

Thank you,

Allen

 

[pongo38]

If you walk up one stair, and you know your weight and the height of the stair, you can work out the work done in achieving this. If someone lifted you up and put you down one step higher, would that change the work done?

 

[PhysicsPete2564]

That makes sense I think, but I feel I'm missing something still. So in the stairs example that force that lifted me was basically overcoming the gravitational potential energy. That gravitational potential energy wouldn't change. But if a force lifted me up wouldn't that in turn change my velocity from 0 to some number? So if my velocity changed shouldn't there be kinetic energy? And then the KE in the stairs example would be force done by someone lifting me up minus the GPE.

I think this is where my thinking may be wrong: In your staris example the force done by someone lifting me would some amount. Then if you subtract GPE from that amount your left with KE. So soundless the force of someone lifting me (or pushing a block up a hill) be more than the GPE because it is changing the velocity?

 

[Mark44]

That makes sense I think, but I feel I'm missing something still. So in the stairs example that force that lifted me was basically overcoming the gravitational potential energy.

The lifting force was overcoming gravity.

That gravitational potential energy wouldn't change.

Sure it would. If you lift something, its potential energy increases.

But if a force lifted me up wouldn't that in turn change my velocity from 0 to some number?

Not necessarily, if the force doing the lifting was equal in magnitude to the frictional force opposing the motion. If you push a heavy box across a floor at a constant speed, the friction force counterbalances the force you're applying, so the net force is zero.

So if my velocity changed shouldn't there be kinetic energy? And then the KE in the stairs example would be force done by someone lifting me up minus the GPE.

I think this is where my thinking may be wrong: In your staris example the force done by someone lifting me would some amount. Then if you subtract GPE from that amount your left with KE. So soundless the force of someone lifting me (or pushing a block up a hill) be more than the GPE because it is changing the velocity?

 

[CWatters]

It does seem as if the answer only includes the change in PE. The block could be accelerating. I take it the problem doesn't say anything about constant velocity?