Question

Consider a drop of rain water having mass $1g$ falling from a height of $1km$. It hits the ground with a speed of $50m/s-1$. Take $g$ constant with a value $10m/s^{-2}$. The work done by the $\left(i\right)$ gravitational force and the $\left(ii\right)$ resistive force of air is:

• A. $\left(i\right)-10J\left(ii\right)-8.255J$
• B. $\left(i\right)1.25J\left(ii\right)-8.255J$
• C. $\left(i\right)100J\left(ii\right)8.75J$
• D. $\left(i\right)10J\left(ii\right)-8.75J$

Answer 1

The correct option is D $\left(i\right)10J\left(ii\right)-8.75J$

Workdone by gravitational force is loss of potential energy $=mgh=1\times {10}^{-3}\times 10\times 1000=\underset{–}{10J}$

Loss of potential energy =gain in kinetic energy + workdone by resistive force

Gain in K.E$=\frac{1}{2}m{v}^2=\frac{1}{2}\times {10}^{-3}\times 50\times 50=1.25J$

Workdone by resistive force$=10-1.25=-8.75J$ (Resistive force of air should in opposite direction against gravitational force that's why negative sign)