Question

Two masses are connected by a light string passing over a light, frictionless pulley as in Figure P5.63. The m₁ = 4.75 kg object is released from rest at a point 4.00 m above the floor, where the m₂ = 3.20 kg object rests. Please define all variables in solving

(a) Determine the speed of each object when the two pass each other.

(b) Determine the speed of each object at the moment the 4.75 kg mass hits the floor.

(c) How much higher does the 3.20 kg mass travel after the 4.75 kg mass hits the floor?

Answer 1

a) The net change in potential energy will equal the increase in kinetic energy

PE = KE

(m1 - m2)gh = ½(m1 + m2)v²
v² = (m1 - m2)gh / ½(m1 + m2)
v² = (4.75 - 3.20)9.81(4 - 2) / ½(4.75 + 3.20)
v = 7.65 m/s

b) same equation, different height
v² = (4.75 - 3.20)9.81(4 - 0) / ½(4.75 + 3.20)
v = 15.3 m/s

c) the velocity energy found in (b) gets converted to potential.

PE = KE
mgh = ½mv²
gh = ½v²
h = v²/2g
h = 15.3²/2(9.81)
h = 11.93 m extra height if there is enough length in the string to allow it.