One end of an elastic cord is attached to a block of mass m and the other end is held keeping the cord relaxed and vertical. Relaxed length of the cord is l and there is a mark P at a height 0.8*l from the lower end. If the upper end is slowly raised, the block leaves the ground when the mark P reaches the point where the upper end of the relaxed cord initially was. Expression for the work done by the force applied by the hand is:
(a) \(\frac{mgl}{16}\)
(b) \(\frac{mgl}{8}\)
(c) \(\frac{mgl}{5}\)
(d) \(\frac{mgl}{4}\)
Answer (1)
To solve this problem, we need to understand the mechanics of how the block and cord system behaves when subjected to forces.
First, let's clarify what's happening in the scenario:
The elastic cord has a relaxed length l .
A mark P is placed at a height 0.8 l from the lower end of this cord.
The upper end of the relaxed cord is initially at a certain point, say, point O .
When the upper end of the cord is slowly raised, the block leaves the ground once the mark P reaches point O .
This means that when the mark P touches the initial position of the upper end, the elastic cord has been stretched by 0.2 l because mark P was 0.8 l away from the lower end originally.
The work done by the force applied by the hand is equal to the elastic potential energy stored in the cord just as the block is about to leave the ground. This is because the energy stored in the stretched cord a moment before the block leaves the ground is used to lift the block.
The elastic potential energy U in the cord when stretched by an amount x is given by:
U = 2 1 k x 2
However, since energy is being transferred to lift the block (and not stored as elastic potential energy), we consider the gravitational potential energy gain that can lift the block: m g h .
Since ma r k P m o v in g t o O implies x = 0.2 l , the work done by the hand against gravity results in raising the center of mass of the cord-block system.
Ultimately, the work done by the force applied by the hand equals the gravitational potential energy change the block undergoes due to lifting:
[ W = mgl \times \text{(fraction length over which cord potential energy fully converts to block lift)}
The block leaves the ground when the mark specified ( P ) in between, covering 1/5 th of the cord.
Thus, the work done is:
W = 5 1 m g l = 5 m g l
Therefore, the correct multiple-choice answer is:
(c) 5 m g l
