Laws of Motion 5 Question 6
6. A ship of mass $3 \times 10^{7} \mathrm{~kg}$ initially at rest, is pulled by a force of $5 \times 10^{4} \mathrm{~N}$ through a distance of $3 \mathrm{~m}$. Assuming that the resistance due to water is negligible, the speed of the ship is
(a) $1.5 \mathrm{~m} / \mathrm{s}$
(b) $60 \mathrm{~m} / \mathrm{s}$
(c) $0.1 \mathrm{~m} / \mathrm{s}$
(d) $5 \mathrm{~m} / \mathrm{s}$
(1980, 2M)
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Answer:
Correct Answer: 6. (c)
Solution:
- As, $a=\frac{F}{m}=\frac{5 \times 10^{4}}{3 \times 10^{7}}=\frac{5}{3} \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}$
So, $v=\sqrt{2 a s}=\sqrt{2 \times \frac{5}{3} \times 10^{-3} \times 3}=0.1 \mathrm{~m} / \mathrm{s}$
