I am working on a problem that has to do with work.
I am also assuming that the acceleration due to gravity is $10m/s^2$.
A 15 kg crate is moved along a horizontal floor by a warehouse worker who's pulling on it with a rope that makes a 30 degree angle with the horizontal. The tension in the rope is 69 N and the crate slides a distance of 10 m. How much work is done on the crate by the worker ? (kinetic friction = 0.4)
I found out that the horizontal force exerted by the rope is about 60N and the force exerted by the friction is about 60N in the opposite direction. So, I cannot see how this object was able to move 10m in the first place.
I understand that the net force = 0 doesn't mean that it is at rest, but I don't quite understand the fact that the problem tells you that it moved 10m.
If I could have answers for the following it would really help.
1), Are we assuming that the crate was already moving ?
2), I calculated the work done by the force by the rope to be 600N and that of the friction to be -600N. Intuitively I want to say that the total work done was 0. But if the object moved, then some work must have been done. What am I thinking wrong ?
Thank you.
Best Answer
Looking at the vertical forces on the crate, you must have an equation like $T \sin 30^{\circ} + F = mg$, where $T$ is the tension on the rope and $F$ is the normal reaction from the floor. This gives $F = 15\times 10 - 69 \sin 30^{\circ} = 115.5\,$N.
Now looking at the horizontal forces, you have $T \cos 30^{\circ} \approx 60$ in one direction and frictional force of $\mu F = 0.4 \times 115.5 \approx 46 $ in the other direction. So, there is a net unbalanced force of about $60-46 = 14\,$N which will accelerate the crate at $\frac{14}{15}$ $m/s^2$.
Total work done is then work done against friction $+$ kinetic energy gained by the acceleration, but this will be equal to force in the horizontal direction times displacement anyway, so $60*10 = 600\,$J.