Physics 12 energy/work question

[gdhillon]

A heavy box of mass “n” kg slides 12.0 m along a straight friction less 30 degree incline. If
the box starts from rest at the top of the incline, what is its speed at the bottom. (hint: think
energy)



I used Fd=1/2mv^2 which was msin30(12)=1-2mv^2 and I canceled the masses then came out with velocity=3.5m/s

 

[howie8594]

If you use "Fd" on the left side of your equation, then instead of msin30(12), it should be mgsin30(12), since mgsin30 is the force pulling the box down the slope.

There's a slightly easier way to do this. Since the incline is 30 degrees, then you know the height of the box has to be half of the length of the slope. The slope is 12 m long, so the height is 6 m. So the potential energy the box has is mg(6). Set that equal to the final kinetic energy of the box. Both methods are basically the same, but you skip a step in this one, and I think it's slightly easier