Question

In which of these situations the work done by the force acting on the slab will be greater?

The crane lifts a concrete slab weighing 1 ton to a height of 10 m, and then moves it horizontally on a section of 20 m. In another situation he moved it horizontally to a distance of 30 m, then lifted to a height of 10 m.

Answer 1

Depending on interpretation of the second move, the answer might be that the work done in both was equal or that the work done in the second situation was greater.

Explanation:

Assuming that the slab is 1 imperial ton, or 2000 pounds, we have
Given: `1 "ton", h_1 = 10 m, d_1 = 20 m`
`" "1 "ton", d_2 = 30 m, h_2 = 10 m`

First we need to convert `1` ton to `kg`: `" "1` ton `~~907.184 kg:`

`(1" ton")/1 xx (2000 lb)/(1" ton")xx (16 oz.)/(1 lb.) xx (28.3495 g)/(1 oz) xx (1 kg)/(1000 g) ~~ 907.184 kg`

Lifting that slab requires a force equal and opposite its weight.

`F = "mass kg" xx 9.8 m/s^2 = 907.184 xx 9.8 ~~ 8890.4 N`; `"Note: "1 (kg*m)/s^2 = 1 N`

`"Work "= "Force" xx "displacement" xx costheta`
where `theta` is the angle between the force and the displacement. The result has units of `N * m = "Joule"`.

The `costheta` part of the above formula causes the result to be zero in cases where the force and displacement are perpendicular. Why?
Because `cos90^@ = 0`

Situation 1:

Analyzing the work done in the 2 phases of the move
`W = W_1 + W_2`

`W = F * h_1*cos0^@ + F * d_1*cos90^@ = F(h_1 + 0)`

`W = 8890.4 N*(10 m + 0) = 88,904 J~~89 kJ`

Situation 2:
An assumption about how this move was done is required. Did the crane operator lift the slab just slightly so that the slab was not dragged during the horizontal move? I will at first assume yes.

`F = 907.185 xx 9.8 ~~ 8890.4 N`

`W = W_1 + W_2`

`W = F * d_2*cos90^@ + F * h_2*cos0^@= F(0 + h_1)`

`W = 8890.4 (0 + 10 m) = 88,904 J~~89 kJ`

This is the same result as for situation 1. So with the first assumption, neither method required more work.

But if we now assume that the slab was dragged along the ground, the crane would have had to exert an unknown horizontal force to overcome the friction with the ground. That horizontal force during the horizontal move means that some work was done during the horizontal move that would increase the total work done in situation 2. So with the second assumption, the second method required more work.