Question

A string id used to pull a block of mass m vertically up by a distance h at a constant acceleration $\dfrac{g}{4}$. The work done by the tension in the string is then

A) $ + \dfrac{{3mgh}}{4}$
B) $\dfrac{{ - mgh}}{4}$
C) $\dfrac{{5mgh}}{4}$
D) $mgh$

Answer 1

Hint: The formula of work done can be used to calculate the correct answer for this problem. Tension on the string always acts away from the body. For the stationary body, the tension is equal to the weight of the body.

Formula used:
The formula of the work done is given by ${\text{W}} = {\text{F}} \times {\text{s}}$ where $F$ is the force and $s$ is the displacement.

Complete step by step answer:
As it is given as the mass of the body is m and the acceleration is given as $\dfrac{g}{4}$ and the acceleration due to gravity is given as $g$ so the tension in the string is given by,
$ \Rightarrow {\text{T}} = {\text{mg}} + {\text{ma}}$
$ \Rightarrow {\text{T}} = {\text{m}}\left( {{\text{a}} + {\text{g}}} \right)$
As it is given as the acceleration is given as ${\text{a}} = \dfrac{{\text{g}}}{{\text{4}}}$.
Therefore,
$ \Rightarrow {\text{T}} = {\text{m}}\left( {{\text{a}} + {\text{g}}} \right)$
$ \Rightarrow {\text{T}} = {\text{m}}\left( {\dfrac{g}{4} + {\text{g}}} \right)$
$ \Rightarrow {\text{T}} = {\text{m}} \times \dfrac{{5g}}{4}$
$ \Rightarrow {\text{T}} = \dfrac{{5mg}}{4}$
As the work done is given by ${\text{W}} = {\text{F}} \times {\text{s}}$ where $F$ is the force and $s$ is the displacement.
Therefore the work done by the tension on the string is given by,
$ \Rightarrow {\text{W}} = {\text{F}} \times {\text{s}}$
Here the force is the tension and the displacement is equal to h.
$ \Rightarrow {\text{W}} = {\text{T}} \times h$
$ \Rightarrow {\text{W}} = \dfrac{{5mg}}{4} \times h$
$ \Rightarrow {\text{W}} = \dfrac{{5mgh}}{4}$

Therefore the work done the tension on the string is given by ${\text{W}} = \dfrac{{5mgh}}{4}$. So the correct answer is option C.

Note:
It is advisable to remember the formula of work done by a body of mass m and having some displacement s. Also, it is important to remember if the direction of applied force and the direction of motion of the body is not the same then there will be a negative sign which indicates that the applied force and the motion of the body are in opposite direction.