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 2006Laws of Motion
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 direction and velocity of ball at A is v m/s.
Then for $PA, v^2 = 0^2 + 2a \times 0.2$
For AB, $0 = v^2 - 2 \times g x^2$
$\Rightarrow v^{2}=2g \times 2$
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]
$\Rightarrow F=m\left(g+a\right)=0.2\left(11g\right)$
$=22\,N$
Alternate solution :
Using work-energy theorem
$W_{mg}+W_{F}=0$
$\Rightarrow -mg \times2.2+F \times 0.2=0 \Rightarrow F=22\,N$
