Conservation of Energy-Two Blocks Sliding Down an Incline

1. Apr 30, 2014

_N3WTON_

1. The problem statement, all variables and given/known data
Block 1 and Block 2 have the same mass 'm', and are released from the top of two inclined planes of the same height making 30 degree and 60 degree angles with the horizontal direction, respectively. If the coefficient of friction is the same in both cases, which of the blocks is going faster when it reaches the bottom of its respective incline?

2. Relevant equations

mgh = 1/2mv^2

3. The attempt at a solution
Using the above equation, I found that both blocks would have the same speed once they reached the bottom of the incline. However, block 2 would reach the bottom first due to it's PE being converted into KE faster than block 1. However, my instructor informed me that my answer "Both blocks have the same speed at the bottom" is incorrect and in fact Block 2 is faster at the bottom. Can anyone explain why this is the case? Thanks.

2. Apr 30, 2014

PhanthomJay

Your energy equation neglected the work done by friction. They would have the same speed if there was no friction. That is not the case here.

3. Apr 30, 2014

_N3WTON_

thank you

4. Apr 30, 2014

Rellek

In case you need a refresher for work and energy equations:

$$T_{1} + U_{1} + \int{Fds} = T_{2} + U_{2}$$

Where $T$ and $U$ are your kinetic energy and gravitational potential energy, respectively.

If the force applied to your object is constant, as is the case with friction, work can simply be defined as force multiplied by the distance over which it is implied.

Using trigonometry, how could you compare the sliding distances of the two blocks, keeping in kind that the heights that the blocks start out at are the same?

I suggest drawing out this scenario with the equivalent forces to help you out.