A body moves in a vertical circle of radius r from the lowest point to highest point. What is the amount of work done by gravitational force during the motion? (Assume mass of the body to be 'm' kg and acceleration due to gravity to be 'g' $\frac{m}{s^{2}}$ )

2mgr

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mgr

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mg $π$ r

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2mg $π$ r

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Solution

The correct option is A
2mgr

Mass of car A, $m_{A} = 5 k g$

Initial speed of car A, $u_{A} = 10 m / s$

Mass of car B, $m_{B} = 10 k g$

Initial speed of car B, $u_{B} = 5 m / s$

Time taken before they come to rest, $d = 10 s e c$ .

So final speed =0
Work done by gravitational force = $f o r c e \times d i s p l a c e m e n t$
= $m g \times 2 r = 2 m g r$

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A body moves in a vertical circle of radius r from the lowest point to highest point. What is the amount of work done by gravitational force during the motion? (Assume mass of the body to be 'm' kg and acceleration due to gravity to be 'g' ms2)

The correct option is A 2mgr Mass of car A, mA=5 kg Initial speed of car A, uA=10 m/s Mass of car B, mB=10 kg Initial speed of car B, uB=5 m/s Time taken before they come to rest, d=10 sec. So final speed =0 Work done by gravitational force = force×displacement = mg×2r=2mgr

A body moves in a vertical circle of radius r from the lowest point to highest point. What is the amount of work done by gravitational force during the motion? (Assume mass of the body to be 'm' kg and acceleration due to gravity to be 'g' $\frac{m}{s^{2}}$ )