Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A boat of mass $1000 \mathrm{~kg}$ goes from rest to speed $20.0 \mathrm{~m} / \mathrm{s}$ in $5.0 \mathrm{~s}$. The water exerts a constant drag force and the acceleration of the boat is constant. If the average power required by the boat is $45000 \mathrm{~W}$, then the magnitude of the drag force is
PhysicsWork Power EnergyTS EAMCETTS EAMCET 2022 (20 Jul Shift 2)
Options:
  • A $500 \mathrm{~N}$
  • B $750 \mathrm{~N}$
  • C $250 \mathrm{~N}$
  • D $1000 \mathrm{~N}$
Solution:
2136 Upvotes Verified Answer
The correct answer is: $500 \mathrm{~N}$
$\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{20-0}{5}=4 \mathrm{~m} / \mathrm{s}^2$
$$
\mathrm{P}_{\mathrm{av}}=\frac{\text { Work done }}{\text { time taken }}=\frac{\mathrm{F}_{\text {boat }} \times \mathrm{S}}{\mathrm{t}}
$$
Now, $\mathrm{S}=\frac{1}{2} \times \mathrm{a} \times 5^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{~m}$
$$
\begin{aligned}
& \text { So, } 45000=\frac{\mathrm{F}_{\text {boat }} \times 50}{5} \\
& \Rightarrow \mathrm{F}_{\text {boat }}=4500 \mathrm{~N} \\
& \text { Now, } \mathrm{F}_{\text {net }}=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \text { ma }=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \mathrm{F}_{\text {drag }}=\mathrm{F}_{\text {boat }}-\mathrm{ma} \\
& =4500-4000=500 \mathrm{~N}
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.