Box slides down incline, air resistance does work. Find Velocity.

[EroAlchemist]

Homework Statement

A 38.7g box slides down an incline of height 12.5m. As it slides, air resistance does -3.71J of work. Find the speed at which it reaches the bottom of the incline.

Homework Equations

E kinetic at bottom = E potential at top - Work air resistance
1/2 mv2 - 1/2 mvorig2 = mgh - (-3.71J)

The Attempt at a Solution

1/2(.0387kg)(v2) = (.0387kg)(9.8m/s2)(12.5m) - (-3.71J)

v= sqrt 8.45J / .01935kg = 20.9m/s

Book gives the correct answer as being 7.30m/s.
Can anybody tell where I went wrong here? Thanks much!

 

[cepheid]

You've basically accounted for the negative nature of the work TWICE. First in your set up for the formula:

E kinetic at bottom = E potential at top - Work air resistance

and then again a second time when you actually plugged in the number (-3.71 J). Only one of these negative signs should be there. A more general way to do it would be to set up the formula without assuming anything about the sign of the work:

E kinetic at bottom = E potential at top + Work

This is a general formula since TWO things contribute (or "add") to the KE at the bottom 1. All of the PE is converted into KE and 2. Any work done on the object during its journey also results in a change in KE.

Now, when you plug a negative number into the equation, it takes into account that the contribution to the KE described by 2 happens to be negative in this example

Intuitively this should make sense as well, since the final KE is smaller than it would have been without the air resistance (whereas in your example, the final KE was larger than it would have been w/o the air resistance, which makes no sense).

 

[EroAlchemist]

Thanks for the help! I added the work this time and got the correct answer.