Question

A hanging weight, with a mass of m1 = 0.375 kg, is attached by a cord to a block with mass m2 = 0.825 kg as shown in the figure below. The cord goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the cord does not stretch and does not slip on the pulley, and that the mass of the cord is negligible. (a) Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown. (b) What is the angular speed of the pulley (in rad/s) after the block has moved this distance?

Answer 1

To find the speed of the block after it has moved a distance of 0.700 m, we can consider the conservation of mechanical energy. The equation relating the speed of the block to the height of the weight is derived using energy methods. The angular speed of the pulley can be found using the equation that relates it to the speed of the block.

Step-by-step explanation:

To find the speed of the block after it has moved a distance of 0.700 m, we can consider the conservation of mechanical energy. Initially, the block is at rest, so it has no kinetic energy. The gravitational potential energy of the hanging weight is given by m1gh, where m1 is the mass of the hanging weight, g is the acceleration due to gravity, and h is the height at which the weight is released. When the block has moved a distance of 0.700 m, the height of the weight is h' = R2θ, where R2 is the outer radius of the pulley and θ is the angle through which the pulley has rotated. The speed of the block is related to the height of the weight by the equation:

m1gh = (1/2)(m1 + m2)v^2 + (1/2)Iω^2 + (1/2)MR2ω^2 + (1/2)k(m2d)^2

where v is the speed of the block, I is the moment of inertia of the pulley, ω is the angular velocity of the pulley, M is the mass of the pulley, k is the coefficient of kinetic friction between the block and the table, and d is the distance the block has moved. From this equation, we can solve for v.

To find the angular speed of the pulley, we can use the equation:

ω = v/R2

where R2 is the outer radius of the pulley and v is the speed of the block.