JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Work, Power & Energy

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A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in

m / s

V = N /10

. Then N is

Answer: 4

Solution

By work-energy theorem:

- \frac{1}{2} k x^{2} - μ m g x = - \frac{1}{2} m V^{2}

. Substituting values:

- \frac{1}{2} (2) (0.06)^{2} - 0.1 (0.18) (10) (0.06) = - \frac{1}{2} (0.18) V^{2}

. Solving:

0.0036 + 0.0108 = 0.09 V^{2} \Rightarrow V^{2} = 0.16 \Rightarrow V = 0.4

m/s

= \frac{4}{10}

m/s. Therefore N = 4.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer

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