Question

A hook of radius 2m weighs 100kg. It rolls along a horizontal floor so that its centre if mass has a speed of 20cm/s . How much work has to be done to stop it

Answer 1

Explanation:

what device is used to measure p.d ?

Answer 2

Answer:

Explanation:

Total energy of the hoop = Translational Kinetic energy + Rotational kinetic energy = $\frac{1}{2}$mv² + $\frac{1}{2}$Iω²

For hook I = mr²                                            

Also v = rω

Thus total energy =  $\frac{1}{2}$mv² +  $\frac{1}{2}$mr²($\frac{v}{r}$)²

Hence the work required to stop the the hook =

its total energy = mv² = 100 x (0.2)² = 4J