A) 50 J
B) 30 J
C) 20 J
D) 10 J
Correct Answer: C
Solution :
Key Idea: Velocity at bottom is \[\sqrt{5}\] times the velocity at top. The energy possessed by a body due to velocity v is given by Difference in \[KE={{K}_{A}}-{{K}_{B}}\] Given, \[{{K}_{A}}=\frac{1}{2}mv_{A}^{2}=\frac{1}{2}m(5\,g\,r)\] \[{{K}_{B}}=\frac{1}{2}mv_{B}^{2}=\frac{1}{2}m(g\,r)\] \[\therefore \] \[\Delta \,KE=\frac{1}{2}m(5\,g\,r)-\frac{1}{2}m(g\,r)=2m\,g\,r\] Given,\[m=1\,kg,\,r=1g=10\,m/{{s}^{2}}\] \[\therefore \] \[\Delta \Kappa \Epsilon =2\times 1\times 10\times 1=20\,J\]You need to login to perform this action.
You will be redirected in
3 sec