A) 80 J
B) -180 J
C) 100 J
D) +180 J
Correct Answer: B
Solution :
As the particle is slow shifted, its kinetic energy remains zero. The total work done on the particle is thus zero. The work done by the external agent should be negative of the work done by gravitational field. The work done by the field is \[\int_{i}^{f}{F\cdot \,\,dr}\] Consider figure. Suppose the particle is taken from O to A and then from A to B. The force on the particle is \[F=mE=(2kg)\,(10N/kg)\,(i+j)\] \[=(20N)\,(i+j)\]You need to login to perform this action.
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