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A particle of mass `2kg` is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The force time graph is shown in figure. Find the velocity of the particle at the end of `10s`. |
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Answer» Using impulse =Change in linear momentum (or area under `F-t` graph) We have, `m(v_f-v_i)=Area` or `2(v_f-0)=1/2xx2xx10+2xx10+1/2xx2xx(10+20)+1/2xx4xx20` `=10+20+30+40` or `2v_f=100` `:. v_f=50m//s` |
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