A force acts on a body of mass $10 \mathrm{~kg}$, resulting in its displacement given as $x=\left(\frac{t^3}{25}\right) \mathrm{m}$, where $t$ is the time in seconds. The work done by the force in $1 \mathrm{~s}$ is

• A

  620 J
• B

  333 J
• C

  524 J
• Correct answer
  D

  60 J

Solution

Given, displacement, $x=\frac{t^3}{25}$ So, velocity, $v=\frac{d x}{d t}=\frac{3 t^2}{25} \mathrm{~ms}^{-1}$ Acceleration, $a=\frac{d v}{d t}=\frac{6 t}{25} \mathrm{~ms}^{-2}$ Force $=$ Mass $\times$ Acceleration $F=10 \cdot\left(\frac{6 t}{25}\right)=\frac{12 t}{5} \mathrm{~N}$ Also, $\quad \frac{d x}{d t}=\frac{3 t^2}{25} \Rightarrow d x=\frac{3 t^2}{25} \cdot d t$ Work done in displacing object by displacement $d x$ is So, total work done by force in first $5 \mathrm{~s}$ is
So, total work done by force in first $5 \mathrm{~s}$ is
$=\frac{36}{5 \times 25} \times\left[\frac{t^4}{4}\right]_0^5=\frac{36 \times(5)^4}{5 \times 25 \times 4}=9 \times 5=45 \mathrm{~J}$

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