Electric Power #P# is given by the expression #P=VxxI#, in the given problem
#=5xx3=15W#
Mechanical Power#="Work done"-:"Time" # Also #"Energy"-:"Time"#
Since Work done #=vecFcdot vecs# , assuming #angle theta# between the Force and displacement vectors be # =0#. Hence #costheta=1#,
#:.# Mechanical Power#="Force"xx"Displacement"-:"Time" #
Also #"Force"="mass"xx"acceleration"#
Since Electric power = mechanical power,
#=>15="mass"xx"acceleration"xx"Displacement"xx1/"Time" # .....(1)
For an constant acceleration of an object we know that
#v^2-u^2=2as# .....(2)
where #v,u,a and s# are final velocity, initial velocity, acceleration and displacement respectively.
Given # u=0#, (2) reduces to
#v^2=2as#
#=>s=v^2/(2a)#
Inserting given values and this value of #s# in equation (1) we obtain
#15=mxxaxxv^2/(2a)xx1/t #, where #t# is time taken
#15=cancel6^3xxcancelaxx((8/3)^2)/(cancel2cancel(a))xx1/t#, solving for #t#
#t=3/15xx(8/3)^2=1.4dot 2 s#
