In Figure (i) the man walks 2 m carrying a mass of 15 kg on his hands. In Figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
Solution 1
In the second case
Case (i)
Mass, m = 15 kg
Displacement, s = 2 m
Work done `W = Fs cos theta`
Where theta = Angle between force and displacement.
`=mgscostheta = 15 xx 2 xx 9.8 cos 90^theta`
= 0 (∵ `cos 90^@ = 0`)
Case (ii)
Mass, m = 15 kg
Displacement, s = 2 m
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.
Therefore, the angle between them, θ = 0°
Since cos 0° = 1
Work done, W = Fs cosθ = mgs
= 15 × 9.8 × 2 = 294 J
Hence, more work is done in the second case.
Solution 2
In Fig. (i), force is applied on the mass, by the man in vertically upward direction but distance is moved along the horizontal.
θ = 90°. W = Fs cos 90° = zero
In Fig. (ii), force is applied along the horizontal and the distance moved is also along the horizontal.
Therefore, θ = 0°.
W = Fs cos θ = mg x s cos 0°
W = 15 x 9.8 x 2 x 1 = 294 joule.
Thus, work done in (ii) case is greater.
