InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. `1600 J`B. `160 J`C. `16 J`D. `1.6 J` |
|
Answer» Correct Answer - C `v=(dx)/(dt) =t^(2)` `W =1/2mv^(2) =1/2mt^(4)` `1/2 xx 2 xx (2)^(4) =16 J` |
|