Finding Work Done by Gravity on Inclined Crate

[bob tran]

Homework Statement

In the figure, a 700-kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. As the force pushes the crate a distance of 3.00 m up the incline, the speed changes from 1.40 m/s to 2.50 m/s. How much work does gravity do on the crate during this process?

Homework Equations

$W=KE_f - KE_i\\ W=Fd\cos{\theta}$

The Attempt at a Solution

$$W=KE_f - KE_i W_{total}=\frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\\ W_{total}=\frac{1}{2}m(v^2_f-v^2_i)\\ W_{total}=\frac{1}{2}(700)(2.5^2-1.4^2)\\ W_{total}=1501.5 \ \texttt{J}\\ \ \\ W=Pd\cos{\theta}\\ W=5600(3)\cos{30}\\ W=14549.2 \ \texttt{J}\\ \ \\ W_{total}=W_g+W\\ W_g=W_{total}-W\\ W_g=1501.5-14549.2\\ W_g=-13047.7 \ \texttt{J}$$
The correct answer is $-10300 \ \texttt{J}$. I am not sure how I would incorporate friction (if at all).

 

[haruspex]

You are making it much too complicated.
How do you usually calculate work done against gravity?

 

[gneill]

They're looking for the work done by gravity. What's the force of gravity on the crate? Sketch in the gravitational force vector on your diagram. What distance does the crate move along the direction of the gravitational force vector?Edit: Oops! haruspex got there first!

 

[bob tran]

$$W=-mgh\\ W=-mg(dcos\theta)\\ W=-(700)(9.8)(3\sin{30})\\ W=-10290 \ \texttt{J or } -10300 \ \texttt{J}$$
Wow. I guess I was thrown off because they mentioned P. Thanks!

Out of curiosity, how would I find the work that friction does?

 

[gneill]

Out of curiosity, how would I find the work that friction does?

Compare the actual results given (change in KE of the crate) with what it would have been if there were no friction. That means sorting out all the works done by the known forces.

 

[bob tran]

Compare the actual results given (change in KE of the crate) with what it would have been if there were no friction. That means sorting out all the works done by the known forces.

So would it be like this?
$$W_f=mgd\sin{\theta}+\frac{1}{2}m(v^2_f-v^2_i)-Pd\cos{\theta}$$

 

[gneill]

That looks promising. You might want to ponder on the signs of the terms. Consider, for example, that the work done by friction should be negative. And the force P is adding energy to the crate while gravity is stealing it.