InterviewSolution
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InterviewSolution
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A force act on a `30 gm` particle as a friction of the particle as a function as given by `x = 3 t - 4 t^(2) + t^(3)`, where `x` is in metros and `t` is in seconds. The work done during the first `4` second isA. 5.28 JB. 450 mJC. 490 mJD. 530 mJ |
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Answer» Correct Answer - A `x=3t-4t^2+t^3` `v=(dx)/(dt)=3-8t+3t^2` `t=0`, `v_1=3(m)/(s)` `t=4s`,`v_2=3-8(4)+3(4)^2` `=3-32+48=19(m)/(s)` `W_(1rarr2)=K_2-K_1=(1)/(2)m(v_2^2-v_1^2)` `=(1)/(2)xx(30)/(1000)(19^2-3^2)` `=0.015xx22xx16=5.28J` |
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