Question

PLEASE HELP!!!!!!!!!! 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 GIVE WHOLE SOLUTION

Answer 1

Answer:

the answer is (m2-m1 )gv

Answer 2

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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