Ravi lifts a dumbbell weighing 5 kg to a height of 2 m from the ground. Calculate the work done by him against the force of gravity. (take g = 10 m/s²)

Question

Question

Answer 1

CONCEPT:

Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:

\(W = \vec F \cdot \vec s\)

Potential energy (PE): The energy possessed by a body by virtue of its position or configuration is called potential energy.
- When an object is lifted from the ground, work is done against the force of gravity, since the force of gravity always acts downwards.

\(PE = mgh\)

Where, m = mass of the body, g = acceleration due to gravity and h = height of the body

CALCULATION:

Given that:

Mass (m) = 5 kg

Height (h) = 2 m

Here the work done by Ravi will get converted into the potential energy of the dumbbell at that height.

Work done (W) = m x g x h

⇒ W = 5 x 10 x 2

⇒ W = 100 J

Hence option 4 is correct.