Question

Calculate the work done by a person in lifting a load of \[20kg\] from the ground and floating it \[1m\] high on a table.

Answer 1

Hint:Work done on an object is equal to change in the energy of that object. In the given question, the person is doing work on the load, hence the work done will be equal to the increase in energy of the load. Since the load isn’t moving, there’s no change in the kinetic energy. Hence, the work done will be entirely due to a change in potential energy.

Complete step-by-step solution:
Mass of the load, \[m = 20kg\]
Height gained by the load, \[h = 1m\]
Acceleration due to gravity, \[g = 9.8m/{s^2}\] (standard value at earth’s surface)
Work done by the person against gravity, \[W = mgh = 20 \times 9.8 \times 1 = 196J\]

Additional Information:
Work done against gravity goes around by another name, gravitational potential energy. So if an object of mass m is raised through a height h, the work done on it will be \[mgh\] , and this amount of energy will be stored in the body as gravitational potential energy. If the height of the body increases, the work done on it is said to be positive whereas if the height decreases, the work done would be negative.

Note:- One common point of confusion is the determination of the value of h. The height to which the object is lifted is not an absolute measurement but a relative one. This means that \[W = mg\vartriangle h\] where \[\vartriangle h\] is simply the difference between the final and the initial heights where the object is kept and not the distance of the final height from the ground.