JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A spring of unstretched length \(l\) has a mass \(m\) with one end fixed to a rigid support .Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity \(v\) is
- A \(\frac {1}{2}\,mv^2\)
- B \(mv^2\)
- C \(\frac {1}{3}\,mv^2\)
- D \(\frac {1}{6}\,mv^2\)
Answer & Solution
Correct Answer
(D) \(\frac {1}{6}\,mv^2\)
Step-by-step Solution
Detailed explanation
We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral: \(T=\int_{m} \frac{1}{2} u^{2} d m\) since the spring is uniform, \(d m=\left(\frac{d y}{L}\right) m,\)…
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