Q138 / 180 · COMEDK 2020 · Physics
A force of $20 \mathrm{~N}$ is applied on a body of mass $5 \mathrm{~kg}$ resting on a horizontal plane. The body gains a kinetic energy of $10 \mathrm{~J}$ after it moves a distance $2 \mathrm{~m}$. The frictional force is
- A$10 \mathrm{~N}$
- Correct answerB$15 \mathrm{~N}$
- C$20 \mathrm{~N}$
- D$30 \mathrm{~N}$
Solution
Given, force $F=20 \mathrm{~N}$
$$
m=5 \mathrm{~kg}
$$
Net force on the body, $$ F_{\text {net }}=F-f_{s} \quad \text { [where, } \int_{s}=\text { friction] } $$ $\therefore$ Work done by displacing $2 \mathrm{~m}$ is given as $$ \begin{aligned} W &=F_{\text {net }} \times \text { displacement } \\ &=\left(F-f_{s}\right) 2 \\ &=\left(20-f_{s}\right) 2 \\ \Rightarrow \quad W &=40-2 f_{s} \end{aligned} $$ According to work-energy theorem, Work done = gain in kinetic energy $$ \begin{aligned} \Rightarrow & 40-2 f_{s} &=10 \\ \Rightarrow & 2 f_{s}=40-10 &=30 \\ \Rightarrow & f_{s} &=\frac{30}{2} \\ &=15 \mathrm{~N} \end{aligned} $$
Net force on the body, $$ F_{\text {net }}=F-f_{s} \quad \text { [where, } \int_{s}=\text { friction] } $$ $\therefore$ Work done by displacing $2 \mathrm{~m}$ is given as $$ \begin{aligned} W &=F_{\text {net }} \times \text { displacement } \\ &=\left(F-f_{s}\right) 2 \\ &=\left(20-f_{s}\right) 2 \\ \Rightarrow \quad W &=40-2 f_{s} \end{aligned} $$ According to work-energy theorem, Work done = gain in kinetic energy $$ \begin{aligned} \Rightarrow & 40-2 f_{s} &=10 \\ \Rightarrow & 2 f_{s}=40-10 &=30 \\ \Rightarrow & f_{s} &=\frac{30}{2} \\ &=15 \mathrm{~N} \end{aligned} $$
Hence, the correct answer is option: 'B'.