Question

e-Techno Classwork IX - Physics (Voi - 1) JEE MAIN & ADVANCED LEVEL-4 & 5 Single Correct Choice Type: 17. A block of mass 4 kg slides on a horizontal frictionless surface with a speed of 2 m/s. It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 N/m, by how much the spring will be compressed?​

Answer 1

Answer:

The spring compress is 0.2 m.

Explanation:

Given that,

Mass of block = 4 kg

Speed = 2 m/s

Force constant = 400 N/m

We need to calculate the distance

Kinetic energy of block is converted into elastic potential energy in spring

Relation between kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2 </p><p>2</p><p>1$

$mv </p><p>2</p><p> = </p><p>2</p><p>1</p><p> </p><p> kx </p><p>2$

$</p><p>x^2=\dfrac{mv^2}{k}x </p><p>2</p><p> = </p><p>k</p><p>mv </p><p>2$

put the value into the formula

$x^2=\dfrac{4\times2^2}{400}x </p><p>2</p><p> = </p><p>400</p><p>4×2 </p><p>2$

$x=\sqrt{0.04}x= </p><p>0.04</p><p>$

$x=0.2\ mx=0.2 m$

Hence, The spring compress is 0.2 m.