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* cos*θ* = *m*g*s*

= 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 = F_{s} 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 = F_{s} 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.