# A mass on an incline

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1. Jan 25, 2016

### b100c

1. The problem statement, all variables and given/known data
Attached

2. Relevant equations
kinetic energy = (1/2) m v^2
potential energy = mgh

3. The attempt at a solution
Did I do this correctly,

At the top, kinetic energy is 0 since it starts at rest. At the bottom we choose the potential to be zero
So using conservation,
$$mgz= \frac{1}{2}m\vec{v_f}^2$$
Then, substitute $$z=s(t)sin\theta$$,
And $$\vec{v_f} = \vec{a}t$$
where $$\vec{a} = \frac{\vec{F}}{m} = \frac{mgsin\theta}{m} = gsin\theta$$

To get $$mgs(t)sin\theta = \frac{1}{2}m(gtsin\theta)^2$$
solve for $$s(t) = \frac{gt^2sin\theta}{2}$$

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2. Jan 25, 2016

### cnh1995

Looks correct to me! You can verify it by putting θ=90° and θ=0°. When θ=90°, it's a free fall, so, s(t)=½gt2, the 2nd kinematical equation with u=0. With θ=0, the body will not move at all, so, s=0.