Question

Find the work done in lifting $5$ cement bags, each of mass $25\text{kg}$ to the top of a building of height $50\text{m}$.(consider $g=10m/s^2$)

• A. $12.5\text{kJ}$
• B. $-25\text{kJ}$
• C. $62.5\text{kJ}$
• D. $-62.5\text{kJ}$

Answer 1

The correct option is C $62.5\text{kJ}$

Given, mass of each cement bag $\left(m\right)=25\text{kg}$
No. of bags $=5$
$\therefore$ Total mass $\left(M\right)=nm=5\times 25=125\text{kg}$
Height of building $\left(h\right)=50\text{m}$
By the FBD of the combined bags:


Force applied to lift the body $\left(F\right)=mg$
Since, applied force and displacement are in the same direction
Amount of work done $\left(W\right)=Fs=mgh$
$=125\times 10\times 50$
$=62500\text{J}$ (or) $62.5\text{kJ}$