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JEE Advanced · Physics · 6. Work Power Energy

A block of mass \(0.18 \mathrm{~kg}\) is attached to a spring of force constant \(2 \mathrm{~N} / \mathrm{m}\). The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of \(0.06 \mathrm{~m}\) and comes to rest for the first time. The initial velocity of the block in \(\mathrm{m} / \mathrm{s}\) is \(v=\frac{N}{10}\). Then, \(N\) is

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

Decrease in mechanical energy
\(=\) Work done against friction
\(\therefore \frac{1}{2} m v^2-\frac{1}{2} k x^2=\mu m g x\)
or \(v=\sqrt{\frac{2 \mu m g x+k x^2}{m}}\)
Substituting the values, we get
\(
v=0.4 \mathrm{~ms}^{-1}=\left(\frac{4}{10}\right) \mathrm{ms}^{-1}
\)
\(\therefore\) Answer is 4 .
Analysis of Question
(i) Question is simple.
(ii) If \(\mu_s\) and \(\mu_k\) two values of coefficient of friction are given, then we will take \(\mu_k\).
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