A) 570 mJ
B) 450 mJ
C) 490 mJ
D) 528 mJ
Correct Answer: D
Solution :
Key Idea: Work done during the first 4s is equal to gain in kinetic energy. We have given, \[x=3t-4{{t}^{2}}+{{t}^{3}}\] So, velocity \[v=\frac{dx}{dt}=3-8\,t+3\,{{t}^{2}}\] At t = 0, \[{{v}_{1}}=3-0+0=3\,m/s\] At t = 4 s, \[{{v}_{2}}=3-8\times 4+3\times {{4}^{2}}\] \[=3-32+48=19\,m/s\] Now work done during t = 0 and t = 4s = gain in kinetic energy \[=\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\] \[=\frac{1}{2}\times 3\times {{10}^{-3}}\,[{{(19)}^{2}}-{{(3)}^{2}}]\] \[[\text{Using}\,\,{{a}^{2}}-{{b}^{2}}=(a+b)\,(a-b)]\] \[=1.5\times {{10}^{-3}}\times [\,(19+3)\,(19-3)]\] \[=1.5\times {{10}^{-3}}\times 22\times 16\] \[=528\times {{10}^{-3}}\,J\] \[=528\,mJ\]You need to login to perform this action.
You will be redirected in
3 sec