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So the question is:

There is a block of mass $m$ travelling along an incline that makes an angle $\theta$ with the horizontal. If the block is pushed up the incline with an initial velocity $v_o$, what is its speed when it crosses a point $x$ meters from where it starts?

By intuition, since the incline if frictionless, the block will rise to height $x_{max}$ before falling back down to $x = 0$, at which point it will have the same velocity as it started with, just in the opposite direction. Using a kinematic equation, I get a final speed of the block of:

$$V_f = \sqrt{2g \sin(\theta)\Delta d+V^2_i}$$

So that is what I would like to derive using the equation $\Delta W = \Delta K + \Delta U$ where $K$ is the kinetic energy and $U$ is the potential energy.

Drawing a free body diagram, I get a force $F=mg \sin(\theta)$, which is the gravitational force and $\Delta W = mg \sin(\theta) d$ where $d$ is the distance traveled factoring in the extra height gained from the initial push which I found using kinematic equations.

This gives $$mg \sin(\theta) d = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i + mgh_f - mgh_i,$$ where $mgh_f = 0$ by the reference point.

$$mg \sin(\theta) d + mgh_f = \frac{1}{2}mv^2_f $$ where $mg \sin(\theta) d = mgh_f$

$$2mg \sin(\theta) d = \frac{1}{2}mv^2_f$$

$$V_f = \sqrt{4g \sin(\theta) \Delta d}.$$

Can anyone tell me why there is a $4$ there instead of a $2$? Also, if someone could show me how to include the intial velocity in this equation and not have to calculate the total distance before hand that would be awesome.

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In your first approach I believe you got a wrong sign. It should be $$v_f = \sqrt{v_0^2 - 2g d \sin(\theta)},$$ otherwise your speed will increase with distance.

In your second approach seems like you are equating the work of the gravitational force to the difference of the mechanical energy. However, the work of a force is equal to the difference in the kinetic energy only, not mechanical energy. Don't forget that weight is doing negative work, since it is directed agaisnt the path. Thus

$$W = - (mg \sin(\theta)) \cdot d = K_f - K_i = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_0^2.$$

Hence

$$-2g d \sin(\theta) = v_f^2 - v_0^2,$$

and therefore

$$v_f = \sqrt{v_0^2 - 2gd \sin(\theta)}.$$

This is the same as before, as it should be.

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  • $\begingroup$ Thank you very much. Regarding your first point, I didn't explain the problem well in that the block has an initial velocity up the incline and we're looking for the speed as it passes a point 1.5 meters below where it started, so the total distance traveled is $x_{max} + 1.5$. $\endgroup$ – user47989 Apr 22 '15 at 2:36

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