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.