The acceleration of the blocks is given by:
a = (Fnet)/(m1 + m2)
Fnet = Fg1 + Fg2 - Ff
Fg1 = m1g
Fg2 = m2g
Ff = μk(m1 + m2)g
where μk is the coefficient of kinetic friction.
Substituting the given values, we get:
a = (m1g + m2g - μk(m1 + m2)g)/(m1 + m2)
a = (1.20 kg * 9.8 m/s2 + 2.90 kg * 9.8 m/s2 - 0.300 * (1.20 kg + 2.90 kg) * 9.8 m/s2)/(1.20 kg + 2.90 kg)
a = 0.837 m/s2
The speed of the blocks after they have moved a distance of 74.0 cm is given by:
v = √(2as)
v = √(2 * 0.837 m/s2 * 0.740 m)
v = 1.45 m/s
Two blocks with masses m1 = 1.20 kg and m2 = 2.90 kg are connected by a massless string, as shown in the Figure. They are released from rest. The coefficent of kinetic friction between the upper block and the surface is 0.300. Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 74.0 cm.
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