QJEEM07Phd506: JEE Main 2007 Physics - Class 11 JEEM07Phd506 Solution: A block slides on a horizontal floor with a speed of It strikes …

Question

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A

2

kg

block slides on a horizontal floor with a speed of

4m/s.

It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is

15N

and spring constant is

10, 000

N/m.

The spring compresses by

Accepted Answer

Let the block compress the spring by

x

before coming to rest.

Initial kinetic energy of the block

=

(potential energy of compressed spring)

+

work done due to friction.

{1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x

10,000{x^2} + 30x - 32 = 0

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

\therefore

x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}

\,\,\,\,\, = 0.055m = 5.5cm.

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