A) 1.8 N-s
B) zero
C) 9 N-s
D) 0.9 N-s
Correct Answer: D
Solution :
[d] Given \[F=600-\left( 2\times {{10}^{5}}t \right)\] The force is zero at time t, given by 0 \[=600-2\times {{10}^{5}}t\] \[\Rightarrow t=\frac{600}{2\times {{10}^{5}}}=3\times {{10}^{-3}}\text{ seconds}\] \[\therefore \operatorname{Impulse}=\int_{0}^{t}{Fdt=\int_{0}^{3\times {{10}^{-3}}}{\left( 600-2\times {{10}^{5}}t \right)dt}}\] \[=\left[ 600t-\frac{2\times {{10}^{5}}{{t}^{2}}}{2} \right]_{0}^{3\times {{10}^{-3}}}\] \[=600\times 3\times {{10}^{-3}}-{{10}^{5}}{{\left( 3\times {{10}^{-3}} \right)}^{2}}\] \[=1.8-0.9=0.9\text{Ns}\]
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