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