MHT CET · Physics · Laws of Motion
If a ship of mass \(4 \times 10^{7} \mathrm{~kg}\) initially at rest is pulled by a force of \(5 \times 10^{4} \mathrm{~N}\) through a distance of \(4 \mathrm{~m}\), then the speed of the ship will be (resistance due to water is negligible)
- A \(5 \mathrm{~ms}^{-1}\)
- B \(1.5 \mathrm{~ms}^{-1}\)
- C \(60 \mathrm{~ms}^{-1}\)
- D \(0.1 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(0.1 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Using, \(F=m a\)
\(5 \times 10^{4} =4 \times 10^{7} \times a \)
\(\Rightarrow \quad a=\frac{5 \times 10^{4}}{4 \times 10^{7}}=\frac{5}{4} \times 10^{-3} \mathrm{~ms}^{-2}\)
Given, \(s=4 \mathrm{~m}, 4=0 \quad \therefore v^{2}=u^{2}+2\) as
or
\(=0+2 \times \frac{5}{4} \times 4 \times 10^{-3}\)
\(v^{2}=10^{-2}\)
or
\(v=0.1 \mathrm{~ms}^{-1}\)
\(5 \times 10^{4} =4 \times 10^{7} \times a \)
\(\Rightarrow \quad a=\frac{5 \times 10^{4}}{4 \times 10^{7}}=\frac{5}{4} \times 10^{-3} \mathrm{~ms}^{-2}\)
Given, \(s=4 \mathrm{~m}, 4=0 \quad \therefore v^{2}=u^{2}+2\) as
or
\(=0+2 \times \frac{5}{4} \times 4 \times 10^{-3}\)
\(v^{2}=10^{-2}\)
or
\(v=0.1 \mathrm{~ms}^{-1}\)
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