JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A ball of mass \(100 \,g\) is dropped from a height \(h =\) \(10\, cm\) on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance \(\frac{ h }{2}\). The spring constant is.......... \(Nm^{-1}\) . (Use \(g=10\, ms ^{-2}\) )
- A \(122\)
- B \(129\)
- C \(127\)
- D \(120\)
Answer & Solution
Correct Answer
(D) \(120\)
Step-by-step Solution
Detailed explanation
By energy conservation \(PE = KE\) \(mg \left( H +\frac{ H }{2}\right)=\frac{1}{2} kx ^{2}\left( x =\frac{ H }{2}\right)\) \(0.100 \times 10 \times \frac{3}{2}(0.10)=\frac{1}{2} k (0.05 \times 0.05)\) \(k =\frac{3 \times 0.10}{0.05 \times 0.05}\)…
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