Question

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/$s^{2}$:

• A. 4 N
• B. 16 N
• C. 20N
• D. 22N

Answer 1

Answer: (D)

22N

Answer 2

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 $W_{mg}+w_{F}=0$ $\Rightarrow-mg\times2.2+F\times0.2=0 \Rightarrow F=22N$