- #1

nothingsus

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## Homework Statement

A constant force F of magnitude 82.0 N is applied to a 3.00 kg shoe box at angle phi = 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?

## Homework Equations

W=Fdcos(theta)

## The Attempt at a Solution

So I understand that you can use net work = W_g + W_f = change in KE = 0 (since constant speed) to solve this so W_f = mgd = 3*9.8*0.150 = , but I'm curious why you can't just use the work equation.

Here's my reasoning. Please explain where I'm incorrect.

The applied force on the box acts at 53 degrees to the vertical and causes a displacement of 0.150m straight up. The angle between the force vector and the displacement vector is 180-53 = 127 degrees. so to calculate the work done on the box by the applied force only

W = 82*0.150*cos(127) = -7.40 J

It's negative (and also the wrong magnitude) which doesn't make sense because energy is supposed to be transferred to the box due to the force to move it upwards...?

Why can't I just use the work equation to find work done by a specific force only (the applied force in this case) like I have in other questions?