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, |
\[{{W}_{mg}}+{{W}_{F}}=0\] |
\[\Rightarrow \] \[-mg\times 2.2+F\times 0.2=0\] \[\Rightarrow \] \[F=22N\] |
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