two bodies of mass m1 and m2 (m2>m1) are connected by a light inextensible string which passes through a smooth fixed pulley. the instantaneous power delivered by an external agent 2 pull m1 with constant velocity v is: PLEASE HELP
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Answer:
(m2 - m1)gv
Explanation:
a = dv/dt
as dv = 0
a = 0
F(net) = 0
T = m2g
T = F + m1g
m2g = F + m1g
F = (m2 - m1)g
P = F.v
= (m2 - m1)gv
Thus the instantaneous velocity is (m2-m1)gv.
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