Ncert Exercises & Solutions

Question 7. 19. A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its center of mass has a speed of 20 cm/s. How much work has to be done to stop it?

r = 2 m
m = 100 kg

v = 20 cm/s = 0.2 m/s
The total energy of the hoop,

E_{T} = \frac{1}{2} {mv}^{2} + \frac{1}{2} {Iω}^{2}

Moment of inertia of the hoop about its center,

I = {mr}^{2}

\begin{array}{l} E_{T} = \frac{1}{2} {mv}^{2} + \frac{1}{2} ({mr}^{2}) ω^{2} \\ \Rightarrow E_{T} = \frac{1}{2} {mv}^{2} + \frac{1}{2} {mr}^{2} ω^{2} (as, v = rω) \\ = \frac{1}{2} {mv}^{2} + \frac{1}{2} {mv}^{2} = {mv}^{2} \end{array}

Required work to be done for stopping the hoop,

W = {mv}^{2} = 100 \times (0 .2)^{2} = 4 J

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