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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
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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, $\mathrm{S}=\frac{1}{2} \times \mathrm{a} \times 5^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{~m}$
$$
\begin{aligned}
& \text { So, } 45000=\frac{\mathrm{F}_{\text {boat }} \times 50}{5} \\
& \Rightarrow \mathrm{F}_{\text {boat }}=4500 \mathrm{~N} \\
& \text { Now, } \mathrm{F}_{\text {net }}=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \text { ma }=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \mathrm{F}_{\text {drag }}=\mathrm{F}_{\text {boat }}-\mathrm{ma} \\
& =4500-4000=500 \mathrm{~N}
\end{aligned}
$$
$$
\mathrm{P}_{\mathrm{av}}=\frac{\text { Work done }}{\text { time taken }}=\frac{\mathrm{F}_{\text {boat }} \times \mathrm{S}}{\mathrm{t}}
$$
Now, $\mathrm{S}=\frac{1}{2} \times \mathrm{a} \times 5^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{~m}$
$$
\begin{aligned}
& \text { So, } 45000=\frac{\mathrm{F}_{\text {boat }} \times 50}{5} \\
& \Rightarrow \mathrm{F}_{\text {boat }}=4500 \mathrm{~N} \\
& \text { Now, } \mathrm{F}_{\text {net }}=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \text { ma }=\mathrm{F}_{\text {boat }}-\mathrm{F}_{\text {drag }} \\
& \Rightarrow \mathrm{F}_{\text {drag }}=\mathrm{F}_{\text {boat }}-\mathrm{ma} \\
& =4500-4000=500 \mathrm{~N}
\end{aligned}
$$
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