A 25 kg block is sliding down a 30 degree incline with an initial velocity of 10 m/s. A force F=150N is applied to the block in the upward direction along the incline. The coefficient of kinetic friction is 0.7. How far down the incline does the block travel before coming to rest?

Question

anonymous · Answer

I'll give you the base equation. Solve for the unknown yourself :)

$$\frac{ 1 }{ 2 }mv _{0}^{2}+mgdsin (\theta) = \mu mgd \cos(\theta) $$

You know every single quantity except d. Solve for it and you should get your answer :)

anonymous · Answer

The simple explanation is that the Kinetic Energy of the block and the work done by gravity on the block must be canceled out by the work down by friction throughout the distance moved. So, you could have thought of it in that way and solve it.