Question

A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity 4ms-1 .it collides with a horizontal Spring of force constant 200 Nm-1. The maximum compression produced in the spring will be

Answer 1

The compression produced in the spring is 0.6 m.

Explanation:

Given that,

Mass = 3 kg

Velocity = 4 m/s

Force constant = 200 N/m

We need to calculate the compression produced in the spring

Using conservation of energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2=\dfrac{1}{2}kx^2$

$\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{mr^2}{2}\times(\dfrac{v}{r})^2=\dfrac{1}{2}kx^2$

$\dfrac{1}{2}mv^2+\dfrac{1}{4}mv^2=\dfrac{1}{2}kx^2$

$x^2=\dfrac{3}{2}\dfrac{mv^2}{k}$

Put the value into the formula

$x^2=\dfrac{3}{2}\times\dfrac{3\times4^2}{200}$

$x=\sqrt{0.36}$

$x=0.6\ m$

Hence, The compression produced in the spring is 0.6 m.

Learn more :

Topic : compression in the string

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Answer 2

this is the answer

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