The block starts from rest as shown in the figure. Find the work done by force of $10 \mathrm{I}$ and friction in the time 0 to $4 \mathrm{~s}$. [Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

• Correct answer
  A

  $240 \mathrm{~J},-96 \mathrm{~J}$
• B

  $250 \mathrm{~J}, 96 \mathrm{~J}$
• C

  $240 \mathrm{~J}, 96 \mathrm{~J}$
• D

  $250 \mathrm{~J},-96 \mathrm{~J}$

Solution

Here, applied force, $F=10 \mathrm{~N}$
Friction force, $f=\mu N=0.2 \times m g=0.2 \times 2 \times 10=4 \mathrm{~N}$ Net force, $F_{\text {net }}=F-f=10-4=6 \mathrm{~N}$ $\therefore$ Acceleration, $a=\frac{F_{\text {net }}}{m}=\frac{6}{2}=3 \mathrm{~ms}^{-2}$ Distance covered in $4 \mathrm{~s}$, $s=0+\frac{1}{2} a t^2=\frac{1}{2} \times 3 \times(4)^2=24 \mathrm{~m}$ Work done by applied force of $10 \mathrm{~N}$, $W_1=F s=10 \times 24=240 \mathrm{~J}$ Work done by friction force of $4 \mathrm{~N}$, $W_2=f s \cos 180^{\circ}=-4 \times 24=-96 \mathrm{~J}$

Hence, the correct answer is option: 'A'.