JEE MAIN - Physics (2006 - No. 42)
A ball of mass $$0.2$$ $$kg$$ is thrown vertically upwards by applying a force by hand. If the hand moves $$0.2$$ $$m$$ while applying the force and the ball goes upto $$2$$ $$m$$ height further, find the magnitude of the force. (consider $$g = 10\,m/{s^2}$$).
$$4N$$
$$16$$ $$N$$
$$20$$ $$N$$
$$22$$ $$N$$
Maelezo
According to energy conservation law,
Work done by the hand and due to gravity = total change in the kinetic energy
Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $$\Delta K$$ = 0
$${W_{hand}} + {W_{gravity}} = \Delta K$$
[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
$$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$$ $$ = 0 \Rightarrow F = 22\,N$$
$$\therefore$$ Option (D) is correct.
Work done by the hand and due to gravity = total change in the kinetic energy
Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $$\Delta K$$ = 0
$${W_{hand}} + {W_{gravity}} = \Delta K$$
[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
$$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$$ $$ = 0 \Rightarrow F = 22\,N$$
$$\therefore$$ Option (D) is correct.