How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
I believe it is $\displaystyle W = Fscos\theta$
$\displaystyle F = mg$
$\displaystyle s = 50$
$\displaystyle cos\theta = cos(42)$ (this works for the horizontal angle)
$\displaystyle W = 24 \times 9.81 \times cos(42) \times 50 = 8748,3J = 8.7kJ$
I think that's right, if you have an answer please post it because there could be another way to do it
Work done is equal to $\displaystyle Fs$, where $\displaystyle F$ is the force acting on the particle and $\displaystyle s$ is the displacement.
If I remember my mechanics correctly, then as the question is asking you the work done by gravity, it is the horizontal displacement that you are after, which is equal to $\displaystyle 50 \sin{48}$.
Then as we are interested in the direction of gravity, would it be $\displaystyle 24g.50 \sin{48}$?