IIT-JEE 2011 · Paper 2 · Physics Q34

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PhysicsInteger+4Question 34 of 60

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 is $V=N/10$ . Then $N$ is

Block on a floor joined by a spring to a wall on the right; an arrow shows the impulse pushing the block towards the wall.

Your answer

Answer and worked solution

Official answer: 4

Energy. All the initial kinetic energy goes into the spring and friction:
$\tfrac12(0.18)V^2=\tfrac12(2)(0.06)^2+0.1(0.18)(10)(0.06),$
i.e. $0.09V^2=0.0036+0.0108=0.0144$ .

Velocity. $V^2=0.16$ , $V=0.4=\frac4{10}$ m/s.

Answer: 4

Topics:Spring potential energy Friction

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IIT-JEE 2011 · Paper 2, Physics question 34