Question

A boy of mass $m$ climbs up a staircase of vertical height $h$ . What would have been the work done if he uses a lift in climbing the same vertical height?
A. $\dfrac{{mg}}{h}$
B. $mh$
C. 0
D. $ - mgh$

Answer 1

Hint: We know that work done is the product of force and displacement in the direction of force. While the man is using a left the direction of displacement will be opposite to the direction of force. Hence, they will be making an angle ${180^ \circ }$. Using this we can calculate the work done.

Complete answer:
It is given that a boy is climbing up the stairs. The vertical height of the stair is given as h.
We need to find the work done if he uses a lift instead of climbing the staircase for the same vertical height.
We know that work done is the product of force and displacement in the direction of force.
We can write in equation form as
$ \Rightarrow $$W = F \bullet S$
$ \Rightarrow W = FS\cos \theta $
Where $F$ is the force, $S$ is the displacement and $\theta $ is the angle made by the direction of force with the direction of displacement.
Now in this case we have the weight of the boy as the force.
That is
$ \Rightarrow $$F = mg$
Since weight is the product of mass and acceleration due to gravity.
The displacement in this case is the vertical height h so
$ \Rightarrow $$S = h$
We know that the force of the weight is acting vertically downward and the lift is moving vertically upward so the displacement here is opposite to the direction of force.
So, the angle between the direction of displacement and the direction of force is ${180^ \circ }$ .
Now let us calculate the work done
On substituting the values in the equation 1, we get
$ \Rightarrow W = mg \times h\cos {180^ \circ }$
$ \Rightarrow W = mgh \times - 1$ [$\because \cos {180^ \circ } = - 1$]
$\therefore W = - mgh$
 This is the work done when the man uses lift for the same vertical height.

So, the correct answer is option D.

Note:
Work can be both positive and negative. If the displacement happens in the direction of force then the work done can be told as positive work. In cases where the displacement is in a direction opposite to the direction of force then the work done will be negative work.