When a solid cylinder is rolling on a level ground at a speed of $4 m s^{- 1}$ . If the kinetic energy is $120 J$ , then the mass of cylinder is:

5 k g

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10 k g

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15 k g

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20 k g

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Solution

The correct option is B $10 k g$
$ω = \frac{v}{R}$ (rolling without slipping)
$\frac{1}{2} (\frac{m R^{2}}{2}) ω^{2} + \frac{1}{2} m V^{2} = 120$
$\frac{1}{4} m V^{2} + \frac{1}{2} m V^{2} = 120$
$m = \frac{4}{3} \times \frac{120}{4^{2}}$
$m = 10 K g$

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When a solid cylinder is rolling on a level ground at a speed of 4ms−1. If the kinetic energy is 120J, then the mass of cylinder is:

The correct option is B 10kgω=vR (rolling without slipping)12(mR22)ω2+12mV2=12014mV2+12mV2=120m=43×12042m=10Kg

When a solid cylinder is rolling on a level ground at a speed of $4 m s^{- 1}$ . If the kinetic energy is $120 J$ , then the mass of cylinder is:

The correct option is B $10 k g$
$ω = \frac{v}{R}$ (rolling without slipping)
$\frac{1}{2} (\frac{m R^{2}}{2}) ω^{2} + \frac{1}{2} m V^{2} = 120$
$\frac{1}{4} m V^{2} + \frac{1}{2} m V^{2} = 120$
$m = \frac{4}{3} \times \frac{120}{4^{2}}$
$m = 10 K g$