Question

A block of mass $m$ is suspended by a light thread from an elevator. The elevator is accelerating upward with a uniform acceleration $a$. The work done by tension on the block in $t$ seconds is $(u=0)$

• A. $\frac{m}{2}ga{t}^2$
• B. $\frac{m}{2}(g-a)a{t}^2$
• C. $0$
• D. $\frac{m}{2}(g+a)a{t}^2$

Answer 1

The correct option is D $\frac{m}{2}(g+a)a{t}^2$

From the ground frame, FBD of block is given as

$T-mg=ma\Rightarrow T=m(a+g)$
Now, displacement of block in the time interval $0\text{to}t$ is,
$S=ut+\frac{1}{2}a{t}^2$
$S=0+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2$
$\therefore$ Work done by tension in the direction of displacement is given as
$W_T=T\cdot S=m(a+g)\times \frac{1}{2}a{t}^2$
$=m(g+a)a{t}^2/2$

Hence, $\left(A\right)$ is the correct answer.