Question

7. A block of mass 4kg 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

W= 1/2 K X^2(work done in spring)

Work done for moving the block of mass is equal to kinetic energy K.E= 1/2 mv^2

so, 1/2 mv^2= 1/2 Kx^2

1/2*4*2*2=1/2*400*x^2

x^2=0.04=0.2

Answer 2

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

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x^2=\dfrac{mv^2}{k}$

put the value into the formula

$x^2=\dfrac{4\times2^2}{400}$

$x=\sqrt{0.04}$

$x=0.2\ m$

Hence, The spring compress is 0.2 m.