Il TEW 6:16 PM 15% C 6] A 15 kg load of bricks hangs from one end of a rope that passes over a small frictionless pulley. A 20 kg counterweight is suspended from the other end of the rope. The system is released from rest. a) What is the acceleration of the system? b) What is the tension in the rope while the load is moving? 15ES 20kg
Il TEW 6:16 PM 15% C 6] A 15 kg load of bricks hangs from one end of a rope that passes over a small frictionless pulley. A 20 kg counterweight is suspended from the other end of the rope. The system is released from rest. a) What is the acceleration of the system? b) What is the tension in the rope while the load is moving? 15ES 20kg
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Il TEW
6:16 PM
15% C
6] A 15 kg load of bricks hangs from one end of a rope that passes over a small
frictionless pulley. A 20 kg counterweight is suspended from the other end of
the rope. The system is released from rest.
a) What is the acceleration of the system?
b) What is the tension in the rope while the load is moving?
15ES
20kg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e830786-058e-49dd-8798-047547f22476%2F0053899b-9d44-448a-8d52-819da268bc6c%2F4r3ybtb.png&w=3840&q=75)
Transcribed Image Text:Il TEW
6:16 PM
15% C
6] A 15 kg load of bricks hangs from one end of a rope that passes over a small
frictionless pulley. A 20 kg counterweight is suspended from the other end of
the rope. The system is released from rest.
a) What is the acceleration of the system?
b) What is the tension in the rope while the load is moving?
15ES
20kg
Expert Solution
Step 1
(a)According to Newton’s second law, the acceleration of an object is directly proportional to the force, and inversely proportional to the mass.
Thu the net force acts on mass m1 is,
The net force acts on mass ,2 is,
Step 2
Equate equation (I) and (II).
Substitute the values in equation (III).
Step by step
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