KCET · Physics · Work Power Energy
A body of mass \(\mathrm{m}\) is travelling with a velocity \(\mathrm{u}\). When a constant retarding force \(\mathrm{F}\) is applied, it comes to rest after travelling a distance \(s_{1}\). If the initial velocity is \(2 \mathrm{u}\), with the same force \(\mathrm{F}\), the distance travelled before it comes to rest is \(s_{2}\). Then,
- A \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
- B \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
- C \(s_{2}=\frac{s_{1}}{2}\)
- D \(s_{2}=s_{1}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Step-by-step Solution
Detailed explanation
By work energy theorem,
and \(\quad \mathrm{Fs}_{2}=\frac{1}{2} \mathrm{~m}(2 \mathrm{x})^{2}\)
From Eqs. (i) and (ii),
\(s_{2}=4 s_{1}\)
and \(\quad \mathrm{Fs}_{2}=\frac{1}{2} \mathrm{~m}(2 \mathrm{x})^{2}\)
From Eqs. (i) and (ii),
\(s_{2}=4 s_{1}\)
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