A) \[128mJ\]
B) \[512mJ\]
C) \[576mJ\]
D) \[144mJ\]
Correct Answer: C
Solution :
[c] here, \[m=4,\,\,g=4\times 1{{0}^{-3}}\,\,kg\] \[\operatorname{x}=4{{t}^{2}}+t\therefore \frac{dx}{dt}=8t+1\frac{{{d}^{2}}x}{d{{t}^{2}}}=8\] Work done. \[W=\int{\operatorname{fdx}}=\int{m\frac{{{d}^{2}}x}{d{{t}^{2}}}\left( \frac{dt}{dt} \right)dt}\] \[=\int\limits_{0}^{2}{\left( 4\times {{10}^{-3}} \right)\left( 8 \right)\left( 8t+1 \right)dt}\] \[=32\times {{10}^{-3}}\int\limits_{0}^{2}{\left( 8t+1 \right)dt =32\times 1{{0}^{-3}}{{\left[ \frac{8{{t}^{2}}}{2}+t \right]}^{2}}_{0}}\] \[=32 \times 1{{0}^{-3}} \left[ 4{{\left( 2 \right)}^{2}}+2-0 \right]=576mJ\]
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