A crate of mass m1 on a frictionless inclined plane is attached to another crate of mass m2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspended in air. The plane is inclined at an angle θ = 36.9°. USE CONSERVATION OF ENERGY to find how fast crate m2 is moving after m1 has travelled a distance of 1.4 m along the incline, starting from rest. The mass of m1 is 12.4 kg and the mass of m2 is 16.3 kg.
(Plzz help me with this question and Use conservation of energy, NOT Newton's 2nd law)
(Plzz help me with this question and Use conservation of energy, NOT Newton's 2nd law)
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if m1 moves 1.4 m up the incline, m2 descends 1.4 m
m1 moves vertically up a distance of 1.4 sin 36.9 = 0.84 m
m2 loses an amount of PE = m2 g (1.4)
m1 gains an amount of PE = m1 g (0.84)
the net change in PE = PE gain by m1 - PE loss by m2 = - 121.6J
this must equal the gain in KE
if the rope is taut, both masses will be moving with the same speed, and we have that
1/2(m1+m2) v^2 = 121.6 J
v= Sqrt[2x121.6/(12.4+16.3)] = 2.91m/s
m1 moves vertically up a distance of 1.4 sin 36.9 = 0.84 m
m2 loses an amount of PE = m2 g (1.4)
m1 gains an amount of PE = m1 g (0.84)
the net change in PE = PE gain by m1 - PE loss by m2 = - 121.6J
this must equal the gain in KE
if the rope is taut, both masses will be moving with the same speed, and we have that
1/2(m1+m2) v^2 = 121.6 J
v= Sqrt[2x121.6/(12.4+16.3)] = 2.91m/s