I am a bit confused with how to find work when there is a free body diagram. I am trying to work out this problem, and in it a box is being pulled at a constant speed by a rope at a constant angle above the horizontal. I am given mass, coefficient of kinetic friction, and the angle.I have drawn my free body diagram, and I think it's pretty accurate. I get the following equations, where $P$ is equal to force being pulled:
\begin{align}
\Sigma F_x&=T\cos\theta-f=ma=0 \\
\Sigma F_y &= T \sin \theta + N – mg = ma = 0
\end{align}
So my unknowns are $P$, $T$, and $N$. I know $f=(\text{coefficient of kinetic friction})N$. I also know the distance the box is pulled.
The work I am trying to solve for is the work being done by the man. I solved for $T$ by isolating $N$ in the second equation and plugging it into the first. Then I'm thinking that $T\cos\theta$ would be the force that I would have to multiply by the distance in order to get work, but it isn't giving me the right answer. Is it because the rope is being pulled up at an angle? Do I need to account for that somehow to find the work done by the man?
Best Answer
There is no acceleration in Y direction. If you consider the X and Y axis like this. Maybe this will help. Work done is P*d. Remember the net force will always be zero because there is no acceleration.