Question

What is the magnitude of the net force exerted on the cement?

A building under construction requires building materials to be raised to the upper floors by cranes or elevators. An amount of cement is lifted 83.7 m by a crane, which exerts a force on the cement that is slightly larger than the weight of the cement. If the network done on the cement is 1.48 X 10^3 J, what is the magnitude of the net force exerted on the cement?

Answer 1

See below

Explanation:

`"Net Work"= "Net Force * Distance"`

Therefore:
`"Net Work"/"Distance"= "Net Force"`

So:
`"Net Force"= (1.48*10^3 J)/(83.7m) approx 17.7 N`

Answer 2

The net force is zero during the lift to the upper floors.

Explanation:

`1.48 X 10^3 J` of work is done on the cement while raising it `83.7 m`. The formula for work is

`W = F*d*costheta`

In this case, F is the tension in the cable that does the work, d is the displacement, and because the displacement is in the direction of the tension, `theta` is `0^@`. To find the value of `F`,

`F = W/(d*cos0^@) = (1.48 X 10^3 J)/(83.7 m*1) = 17.7 N`

The question asks for net force - that means the sum of all forces acting on the cement. Cranes lift at constant velocity. Therefore, the cement was not accelerated while being lifted (except very briefly at the beginning and at the end).

The cement also has weight, which is another force acting on the cement.

`F_"net" = |"weight"| - |"tension"|`

Because the acceleration was zero, the net force was zero. The 17.7 N tension in the cable would be equal and opposite the weight of the cement, yielding a net force of zero, while it was lifted at constant speed. That seems to be the answer to your question. Zero net force.

Could this be a trick question?

Notice that the cement has a weight of 17.7 N. That would be less than 2 kg in mass. That is a ridiculously small task to give a crane. This makes me suspect an error in the data provided or as I said above, this might be a trick question.

I hope this helps,
Steve