Q. 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 die force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/$s^{2}$:
AIEEEAIEEE 2008
Solution:
The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B let acceleration of ball during PA is a m/$s^{2}$[assumed to be constant] in upward A direction and velocity of ball at A is v m/s.
Then for PA, $v^{2}=0^{2}+2a\times0.2$
$For AB, \, \quad0=v^{2}-2\times g\times2$
$\Rightarrow \, \quad v^{2}=2g\times2$
From above'equations,
$a=10g=100 m /s^{2} $
Then for PA, FBD of ball is
F - mg = ma[ F is the force exerted by hand on ball]
$\Rightarrow \,\quad$ F = m(g + a) = 0.2(1 lg) = 22N
Alternate solution :
Using work-energy theorem
$W_{mg}+w_{F}=0$
$\Rightarrow -mg\times2.2+F\times0.2=0 \Rightarrow F=22N$
