# Simple net work question

1. Nov 3, 2009

### RoganSarine

Basically my instructor is confusing the heck out of me because I highly dislike how this course is structured. I dislike how in this specific chapter we are not taught:

$$\Sigma$$W = $$\Delta$$Energy

However, using this equation and my previous knowledge I can't seem to get the right answer in the following problem unless I assume there is no $$\Delta$$PE which isn't true since $$\Delta$$height = .150

When using the relationship you see below, do we assume that the Work done by gravity doesn't exist since we compensate for it under the change in potential energy?

1. The problem statement, all variables and given/known data
In Figure 7-32, a constant force of magnitude 82.0 N is applied to a 3.00 kg shoe box at angle f = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by when the box has moved through vertical distance h = 0.150 m??

2. Relevant equations

http://www.lowellphysics.org/beta/T...rk/Problems/c07x7_11.xform_files/nw0314-n.gif

$$\Sigma$$W = $$\Delta$$Energy
$$\Sigma$$W = $$\Delta$$KE + $$\Delta$$PE
Wg + Wapp = 0(constant speed) + $$\Delta$$PE

Wg + Wapp = $$\Delta$$PE

3. The attempt at a solution

Using the above formula:
Wg + Wapp = $$\Delta$$PE

Wg + Wapp = (3)(g)($$\Delta$$height)
Wg + Wapp = (3)(g)(-.15)
Wg + Wapp = -4.4145

Wg = F * d * cos$$\Phi$$
Wg = mg * (.15) * cos(0)
Wg = (3)(9.81) * (.15)
Wg = 4.4145

Wapp = F * d * cos$$\Phi$$
Wapp = 82cos53 * (.15) * cos0
Wapp = 7.4

This relationship doesn't make any sense...
4.41 + 7.4 $$\neq$$ -4.4145

Basically, if I assume that there is no change in potential energy, however, the answer does end up coming out right... kinda of.

Last edited: Nov 3, 2009
2. Nov 3, 2009

### tiny-tim

Hi RoganSarine!

(have a phi: φ and a delta: ∆ and a sigma: ∑ )
Nooo

potential energy is just another name for (minus) work done by a conservative force.

You can either use Wg or use ∆PE, but not both!!