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, \[{{W}_{mg}}+{{W}_{F}}=0\] \[\Rightarrow \] \[-mg\times 2.2+F\times 0.2=0\] \[\Rightarrow \] \[F=22N\]You need to login to perform this action.
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