# Intuitive understanding of work.

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.

• I think you may find better assistance at Physics SE. physics.stackexchange.com Commented Oct 24, 2013 at 19:01
• It is an approximation. If it moves at constant velocity then the forces balance out. (Also, $g=9.81$ m/s is a more typical approximation.) Commented Oct 24, 2013 at 19:03
• If the crate is pulled, have you considered the vertical component of tension reducing the normal force, and hence friction? Commented Oct 24, 2013 at 19:08

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.

• The work done by the worker is $600$ Joules regardless of friction. The net work is simply $14\cdot 10 = 140$ J. No need to calculate any accelerations etc. Commented Oct 24, 2013 at 19:38
• @user7530 Agreed that there is no need to calculate those separately, that was more for better understanding if one needs to check. The work done to overcome friction is 460J, the remaining work goes into kinetic energy (140 J). Commented Oct 24, 2013 at 19:44
• I totally forgot about the vertical component of the tension. Thank you so much ! Commented Oct 24, 2013 at 21:33

If you move an object horizontally, without imparting on it any net change in velocity, then you have done no work on it.

One way to think about this is using conservation of energy: the initial energy of the box, plus the work done on it, must equal the final energy of the box. Since the box has no kinetic energy at the start and end, and the same potential energy, the work must be zero.

However,

1) It seems your calculation for the force of friction is off -- see Macavity's calculations below;

2) the problem asks you for the work done on the box by the worker, not the total work done on the box. Taking on faith your calculation that the horizontal component of tension on the string is 60 N, the work done by the worker on the box is $$60 \textrm{ N} \cdot 10 \textrm{ m} = 600 \textrm{ J}.$$

Notice that this is irrespective of however much work might be done by friction (460 N, it seems).

• @Macavity Yes, I didn't check the OP's calculations. Commented Oct 24, 2013 at 19:49
• Thanks for the help everyone ! Commented Oct 24, 2013 at 21:34