A force of F=10t+3t2 is acting on a particle on a particle of mass 1Kg at rest, where F is in newton and t is in seconds. Work done by this force in first 1 second is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
18 J
b
14 J
c
32 J
d
36 J
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
F = 10t+3t2 mdvdt= 10t+3t2 dv= (10t+3t2)dtv= 10t22+t30 1v after 1 sec 6 m/s. work done is equal to change in K.E= 12(1)(6)2= 18 J