Q129 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics

A force of $4 \mathrm{~N}$ acts on a $10 \mathrm{~kg}$ body initially at rest. Let $W_1$ is work done by force during $0 \leq t \leq$ ls. Likewise $W_2$ is the work done by force during ls $\leq t \leq 2 \mathrm{~s}$, where $t$ is time in second. The ratio $\frac{W_2}{W_1}$ is

  • A
    1
  • B
    2
  • C
    $\frac{1}{2}$
  • Correct answer
    D
    3

Solution

Mass of body, $m=10 \mathrm{~kg}$ Force, $\quad F=4 \mathrm{~N}$ Acceleration produced in the body due to applied force, $a=\frac{F}{m}=\frac{4}{10}=0.4 \mathrm{~ms}^{-2}$ For $0 \leq t \leq 1 \mathrm{~s}, t_1=1 \mathrm{~s}$ $\therefore$ Distance travelled by the body is given as $s_1=u t_1+\frac{1}{2} a t_1^2=0 \times 1+\frac{1}{2} \times 0.4 \times 1^2=0.2 \mathrm{~m}$ $\therefore$ Work done, $W_1=F \times s_1=4 \times 0.2=0.8 \mathrm{~J}$ For $\mathrm{l} \leq t \leq 2 \mathrm{~s}, t_2=1 \mathrm{~s}$ Velocity of the body after $t_1=1 \mathrm{~s}$ is $v$, then $v=0+a t_1=0+0.4 \times 1=0.4 \mathrm{~ms}^{-1}$ Distance travelled by the body in this situation is given as, $s_2=v t_2+\frac{1}{2} a t_2^2=0.4 \times 1+\frac{1}{2} \times 0.4 \times 1^2=0.6 \mathrm{~m}$ $\therefore$ Work done, $W_2=F s_2=4 \times 0.6=2.4 \mathrm{~J}$ $\therefore \quad \frac{W_2}{W_1}=\frac{2.4}{0.8}=\frac{3}{1}=3$

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

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