Question

A porter lifts a heavy suitcase of mass $80\text{kg}$ and at the destination lowers it down by a distance of $80\text{cm}$ with a constant velocity. Calculate the work done by the porter in lowering the suitcase. $(\text{take}g=9.8{\text{ms}}^{–2})$

• A. $–627.2\text{J}$
• B. $-62720.0\text{J}$
• C. $784.0\text{J}$
• D. $+627.2\text{J}$

Answer 1

The correct option is A $–627.2\text{J}$


Let $F$ is the force applied by the porter to lower the suitcase.

$W.D_{porter}=F.d=Fd\cos {180}^{\circ }$

As the object is moving with constant velocity, $F=80g$

$\Rightarrow W.D_{porter}=80g\times 0.8\times \cos {180}^{\circ }$

$\Rightarrow W.D_{porter}=80\times 9.8\times 0.8\times (-1)$

$\Rightarrow W.D_{porter}=627.2\text{J}$


Alternate solution:

From work energy theorem,

$W_{\text{Porter}}+W_{\text{mg}}=\Delta K.E.=0$

$W_{\text{Porter}}=-W_{\text{mg}}=-\text{mgd}$

$=-80\times 9.8\times 0.8=-627.2\text{J}$

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