ExamBro
ExamBro
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

  1. A \(\frac {1}{2}\,mv^2\)
  2. B \(mv^2\)
  3. C \(\frac {1}{3}\,mv^2\)
  4. D \(\frac {1}{6}\,mv^2\)
Verified Solution

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, \(d m=\left(\frac{d y}{L}\right) m,\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app