Two blocks are connected by a massless string across a frictionless pulley. Bloc
ID: 1487770 • Letter: T
Question
Two blocks are connected by a massless string across a frictionless pulley. Block 1 with a mass of m_1 = 10.0 kg sits on a frictionless horizontal surface and block 2 with a mass of m_2 = 15.0 kg hangs off the surface. Block 1 is also connected to a spring with spring constant of k = 1350 N/m. Initially the spring is at its equilibrium position and both blocks are at rest. You release the blocks and they are traveling at a speed of 0.400 m/s when the spring is stretched 0.200 m. If the string across the pulley doesn't slip, what is the mass of the pulley?Explanation / Answer
use conservation of energy remembering there will be rotational energy due to the pulley
m1 moves down 0.75m and m2 moves up 0.75m, this means there is a loss of PE equal to
(m2-m1)gh = 5kg x 9.8m/s/s =49 j
this amount of energy will be distributed among the translational KE of the 10kg and 15 kg masses and also of the pulley
the two masses have combined KE of 1/2(10kg+15kg)(0.4m/s)^2 = 10J, meaning 49 J is in the form of rotational KE of the pulley
the rotational KE of the pulley is 1/2 I w^2
where I is the moment of inertia and is equal to 1/2 MR^2 where M, R are the mass and radius of the pulley, and w = v/R is its angular velocity
we know the tangential velocity is 0.4m/s, so that its angular velocity is
w = 0.4m/s / 0.12m = 3.33rad/s
therefore, 1/2 (1/2 MR^2) w^2 = 1.7J
1/4M(0.4m)^2(15rad/s)^2 = 1.7J
M = 2.1kg