# Math Help - Gravity-Work Question

1. ## Gravity-Work Question

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?

2. Originally Posted by peon123
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 $W = Fscos\theta$

$F = mg$
$s = 50$
$cos\theta = cos(42)$ (this works for the horizontal angle)

$W = 24 \times 9.81 \times cos(42) \times 50 = 8748,3J = 8.7kJ$

I think that's right, if you have an answer please post it because there could be another way to do it

3. Originally Posted by peon123
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?
Work done is equal to $Fs$, where $F$ is the force acting on the particle and $s$ 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 $50 \sin{48}$.

Then as we are interested in the direction of gravity, would it be $24g.50 \sin{48}$?

4. This works out to be $8739.4J$ or $8.7kJ$.

Just to note, I have taken $g$ as 9.8, as opposed to 9.81.

5. Just realised that we used the same equations

Originally Posted by e^(i*pi)
$W = 24 \times 9.81 \times cos(42) \times 50 = 87483J ~ 8.7kJ$
There's a typo in your answer, it should be $8748.3J$, not $87483J$

6. Originally Posted by craig
Just realised that we used the same equations

There's a typo in your answer, it should be $8748.3J$, not $87483J$
Yeah, I checked that on my calculator, makes sense since $sin(90-x) = cos(x)$

Thanks for that, I worked it out in kJ first but had to expand for the 1dp