question_answer

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 up to 2 m height further, find the magnitude of the force. Consider\[g=10\text{ }m/{{s}^{2}}\].     AIEEE  Solved  Paper-2006

A) 4N             

B)                        16N                       

C)        20N                       

D)        22N

Correct Answer: D

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 S, let acceleration of ball during is a\[m/{{s}^{2}}\] (assumed to be constant) in upward direction and velocity of ball at A is\[vm/s\]. Then, for PA,                   \[{{v}^{2}}={{0}^{2}}+2a\times 02\] For AB, \[0={{v}^{2}}-2\times g\times 2\] \[\Rightarrow \] \[{{v}^{2}}=2g\times 2\]           From above equation,           \[a=10\,g=100\,m/{{s}^{2}}\] Then, for PA, FBD of ball is \[F-mg=ma\](F is the force exerted by hand on ball) \[\Rightarrow \] \[F=m(g+a)=02(11g)=22\,N\] Alternate Solution Using work-energy theorem, \[{{W}_{mg}}+{{W}_{F}}=0\] \[\Rightarrow \] \[-mg\times 2.2+F\times 0.2=0\] \[\Rightarrow \] \[F=22N\]