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

A) 4N

B) 16N

C) 20N

D) 22N

Correct Answer: D

Solution :

[d] 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,

\[{{W}_{mg}}+{{W}_{F}}=0\]

\[\Rightarrow \] \[-mg\times 2.2+F\times 0.2=0\] \[\Rightarrow \] \[F=22N\]