Question

a ring of mass 6 kg and radius 2 m rolls along a horizontal floor such that the speed of its centre is 4 m/s. the work to be done to stop the ring is​

Answer 1

Answer:

corect answer is 96 (mv2=6×4×4)

Answer 2

Answer:

The work to be done to stop the ring is 48Joules

Explanation:

Given: Mass(m)=6kg

           Distance(r)=2m

           Velocity(v)=4m/s

To find: Work to be done to stop the ring(W)=?

Solution:
Since the ring is moving along the horizontal floor, the work done will be equal to kinetic energy.

We know that,

$KE=1/2mv^{2}$

      = 1/2*6*4*4

      =48J

Thus, the Work to be done to stop the ring is 48Joules