Question

A weightlifter jerks 220 kg vertically through 1.5 meters and holds still at that height for two minutes. The work done by him in lifting and in holding it still are respectively:
A) 220J, 330J
B) 3234J, 0
C) 2334J, 10J
D) 0, 3234J

Answer 1

The weight lifted by the weight lifter and the height to which it is lifted is given, the work done in doing so can be calculating using the formula of the work. The force is done in the direction opposite to the gravitational pull (acting downwards).
We will be using the value of acceleration due to gravity ‘g’ as $9.8m/{s^2}$

Formula to be used:
$W = FS\cos \theta$ where W is the work done, F is the force, S is the displacement and $\theta$ is the angle formed between the force and the displacement.

Complete step by step answer:
The weight lifter lifts 220 kg weight, in order to do so he has to do work against the gravity.

Work done is given as the dot product of force and displacement vector and dot product changes to $\cos \theta$
$W = \vec F.\vec S \\\ \implies W = FS\cos \theta .....(1) \\\$
i) Work done in lifting the weight by the weightlifter can be given as:
Work done = W (say)
Force applied (F) = mg
$= 220 \times 9.8 \\\ = 2156N \\\$
Displacement (S) = height till where the weight is lifted
= 1.5 m
As both force and displacement are in same direction, the angle between them is 0°
Angle formed $\left( \theta \right)$ = 0 and $\cos 0 = 1$
Substituting the values in (1), we have:
$W = 2156 \times 1.5 \times 1 \\\ \implies W = 3234J \\\$
🡪 Work done by weightlifter in lifting the weight to the given height is 3234 Joules (SI Unit of work)
ii) Work done in holding the weight by the weightlifter at given height:
When the mass is held still at the given position, it doesn’t cover any distance and thus has no displacement because the initial and final positions of the weight are the same. When displacement becomes 0, the work done by equation (1) also becomes 0 as when 0 is multiplied with other numbers, the result is 0 as well.
🡪 Work done by the weightlifter in holding the weight by the weightlifter at a given height is 0.

Therefore, the work done by the weightlifter in lifting and in holding it still are 3234 J and 0 respectively and the correct option is B).

Note:
Dot product among the vectors changes to cos and cross product changes to sine of the angle formed between the two vectors.
Remember that displacement is given by the difference between the final and initial positions of the object and when both the positions are the same, the displacement becomes zero.
The SI unit of work is joules (J) and we always have to give the final answer along with the SI Units.