A light, inextensible cord passes over a light, frictionless pulley with a radiu

Question

A light, inextensible cord passes over a light, frictionless pulley with a radius of 10 cm. It has a(n) 18 kg mass on the left and a(n) 4.3 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 4.1 m apart. The acceleration of gravity is 9.8 m/s 2 . 4.1 m 10 cm 18 kg 4.3 kg At what rate are the two masses accelerating when they pass each other? Answer in units of m/s 2 . 008 (part 2 of 2) 10.0 points What is the tension in the cord when they pass each other? Answer in units of N. 009 (part 1 of 3) 10.0 points A 2.22 kg block slides down a smooth, frictionless plane having an inclination of 30 . The acceleration of gravity is 9.8 m/s 2 . 2.22 kg 2.76 m 30 Find the acceleration of the block. Answer in units of m/s 2 . 010 (part 2 of 3) 10.0 points What is the block’s speed when, starting from rest, it has traveled a distance of 2.76 m along the incline. Answer in units of m/s. 011 (part 3 of 3) 10.0 points What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.76 m down the incline? Answer in units of N.

Explanation / Answer

given pulley radius, r = 0.1 m

m1 = 18 kg

m2 = 4.3 kg

initial seperation, x1 = 4.1 m

a. acceleration = a

from force balance

a = (m1 - m2)*g/(m1 + m2) = 6.0267713 m/s/s

b. tension in cord = T

from force bnalance

T - m2g = m2a

T = 68.09811 N

given m = 2.22 kg

theta = 30 deg

acceleration a = g*sin(theta) = 4.905 m/s/s

now, speed after travelling d = 2.76 m

2*a*d = v^2

v = 5.203421 m/s

normal reaction = mg*cos(theta) = 18.8604744486 N

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