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Grade 11Mechanics

force F is equal to 3ti^ + 5j^ in acts on a body due to which its position varies as S is equal to 2t^2i^ - 5j^. Work done by this force in first 2 seconds is

Profile image of Vivek kumar Ojha
8 Years agoGrade 11
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1 Answer

Profile image of Arun
8 Years ago
Dear Vivek
 
Work done = F. dS
= (3t i + 5j).(2 t² i - 5j)
= 6t³ -25
 
At t= 2 sec
Work done = 6 *2³ -25 = 48- 25 = 23 Joules
 
Regards
Arun (askIITians forum expert)