JEE Main · 2007 · 29 Apr 2007 · Offline · Work, Power & Energy

A 2 kg block slides on a horizontal floor with a speed of

4 m/s

. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is

10, 000 N/m

. The spring compresses by

(A)5.5 cm✓ Correct
(B)2.5 cm
(C)11.0 cm
(D)8.5 cm

Answer: (A)

Solution

Use energy conservation from the moment of contact until the block stops: the initial kinetic energy goes partly into the spring and partly into work done against friction, so

\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2} + f x

. Putting

\frac{1}{2} (2) (4)^{2} = 16

k = 10000

N/m and

f = 15

N gives

5000 x^{2} + 15 x - 16 = 0

. The positive root is about

x \approx 0.055

m, so the spring is compressed by roughly

5.5

cm.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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