Question

a circular ring of radius 2 metre with hundred kg it roles along a horizontal floor so that its centre of mass has a speed of 20 m per second how much work has to be done to stop it.

Answer 1

Answer:

Work done to stop the ring is given as

$W = 4 \times 10^4 J$

Explanation:

As we know by work energy Theorem

Work done by all forces must be equal to change in kinetic energy of the system

so we will have

$\Delta K = W$

$W = \frac{1}{2}I\omega ^2 + \frac{1}{2}mv^2$

$W = \frac{1}{2}(mR^2)(\frac{v^2}{R^2} + \frac{1}{2}mv^2$

$W = mv^2$

$W = (100)(20^2)$

$W = 4 \times 10^4 J$

#Learn

Topic : Rolling motion

https://brainly.in/question/2114572