NEET · Physics · STD 11 - 5. work,energy,power and collision
The kinetic energies of two similar cars \(A\) and \(B\) are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If \(F _{ A }\) and \(F _{ B }\) are the forces applied by the breaks on cars A and B , respectively, then the ratio \(F _{ A } / F _{ B }\) is :
- A \(\frac{3}{2}\)
- B \(\frac{2}{3}\)
- C \(\frac{1}{3}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\because \text {W.D. }=\Delta KE \)
\( \overrightarrow{ F } \cdot \overrightarrow{ S }=\Delta KE \)
\( \frac{(\Delta K E)_A}{(\Delta K E)_B}=\frac{- F _A S_A}{- F _{ B } S _{ B }} \)
\( \frac{-100}{-225}=\frac{- F _{ A }(1000)}{- F _{ B }(1500)} \)
\( \frac{ F _{ A }}{ F _{ B }}=\frac{2}{3}\)
\( \overrightarrow{ F } \cdot \overrightarrow{ S }=\Delta KE \)
\( \frac{(\Delta K E)_A}{(\Delta K E)_B}=\frac{- F _A S_A}{- F _{ B } S _{ B }} \)
\( \frac{-100}{-225}=\frac{- F _{ A }(1000)}{- F _{ B }(1500)} \)
\( \frac{ F _{ A }}{ F _{ B }}=\frac{2}{3}\)
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