Does Ignoring Work Done by Gravity Affect Calculations in Work-Energy Problems?

[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?

Homework Statement

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. Homework 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
W_g + W_app = 0_{(constant speed)} + $$\Delta$$PE

W_g + W_app = $$\Delta$$PE


3. The Attempt at a Solution
Using the above formula:
W_g + W_app = $$\Delta$$PE

W_g + W_app = (3)(g)($$\Delta$$height)
W_g + W_app = (3)(g)(-.15)
W_g + W_app = -4.4145

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

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

 

[tiny-tim]

Hi RoganSarine!

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

W_g + W_app = 0_{(constant speed)} + $$\Delta$$PE

Nooo …

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

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

 

[kerimek]

It seems that your confusion stems from the fact that your instructor has not explicitly taught the relationship between work and energy in this specific chapter. However, the equation \SigmaW = \DeltaEnergy is a fundamental concept in physics and should be applied in all relevant scenarios, including the one presented in this problem.

In this problem, the work done on the box by the applied force (Wapp) and the work done by gravity (Wg) both contribute to the change in the box's total energy. This is represented by the equation \SigmaW = \DeltaKE + \DeltaPE, where \DeltaKE is the change in kinetic energy and \DeltaPE is the change in potential energy. Since the box is moving at constant speed, there is no change in kinetic energy and thus \DeltaKE = 0. This leaves us with the equation Wg + Wapp = \DeltaPE.

In order to calculate the work done by gravity, you correctly used the equation Wg = mgdcos\Phi where m is the mass of the box, g is the gravitational acceleration, d is the vertical distance the box moves, and \Phi is the angle between the force of gravity and the direction of motion. In this case, \Phi = 0 since the box is moving vertically. However, for the work done by the applied force, you seem to have used the wrong formula. The correct formula is Wapp = Fdcos\Phi where F is the magnitude of the applied force and \Phi is the angle between the applied force and the direction of motion. In this case, \Phi = 53° and F = 82 N. Using this formula, you should get Wapp = 82cos53 * (.15) = 7.4 J.

Thus, the correct equation for this problem is Wg + Wapp = \DeltaPE. Substituting the values we calculated, we get 4.4145 J + 7.4 J = \DeltaPE. This gives us a total change in potential energy of 11.8145 J. This is the work done on the box by the applied force and gravity combined, and it is equal to the change in potential energy of the box. I hope this explanation helps clarify the relationship between work and energy in this problem.