Another H.W. problem I'm stumped on. A light inextensible cord passes over a light frictionless pulley?

Question

with a radius of 4.9 cm. A mass on the left is 15 kg, and a mass on the right is 4.9 kg. The masses are hanging freely with the center of their masses a vertical distance of 2.7 m apart. (Acceleration of gravity = 9.8 m/s2)

What is the rate that the two masses are accelerating when they pass each other (answer in m/s2)?

What is the tension in the cor when they pass each other (answer in N)?

Answer 1

Its always best to start with a free-body diagram with these types of problems.

Newton's second law is the key to the problem.

F = ma
Sum of all forces = 0

F1 = T - m1*g = m1*a
F2 = m2*g - T = m2*a

So if you combine the two equations, the T's cancel out and solving for acceleration you get:

a = [(m2 - m1)/(m1 + m2)] * g

Once you have the acceleration you can find the tension using the same technique...

T + mg = a
T = m(a - g)

Answer 2

2 units are appropriate through a worry-free string passing over a worry-free frictionless pulley, awesome from a ceiling. the article of mass 3.00kg is released from resting on the floor. the 2d merchandise of 5.00kg, awesome on the diverse component of the pulley, starts to flow from its unique excellent of four.00 m above the floor. employing the isolated kit variety locate the optimal excellent which the three kg merchandise rises. because the 5 kg merchandise quickens down 4.00 meters, the three kg merchandise quickens up 4.00 meters. Then the three kg merchandise decelerates till ultimately its very very last speed = 0 m/s the web stress on both units = Weight of 5 kg merchandise – Weight of three kg merchandise = 5 * 9.8 – 3 * 9.8 = 19.6 N information superhighway stress = finished mass * acceleration 19.6 = 8 * a a = 19.6 ÷ 8 = 2.40 5 m/s^2 the three kg merchandise quickens at 2.40 5 m/s^2 for a distance of four.00 meters. very very last speed^2 – initial speed^2 = 2 * a * d initial speed = 0 a = 2.40 5 m/s^2 d = 4 m very very last speed^2 = 2 * 2.40 5 * 4 very very last speed = (2 * 2.40 5 * 4)^0.5 = 19.6^0.5 Then the three kg merchandise decelerates at 9.8 m/s till ultimately its speed = 0 m/s. very very last speed^2 – initial speed^2 = 2 * a * d very very last speed^2 = 0 initial speed^2 = 19.6 a = -9.8 0 – 19.6 = 2 * -9.8 * d d = a million meter finished distance the three kg merchandise moved up = 4 + a million = 5 meters the reply is 5.00 m.