A block of mass m1 =3 kg is on a table and attached to a block of mass m2 =4 kg m2 =4 kg by a mass less string. The coefficient of friction is 0.28 between the table and m1. The system is released from rest. After m2 falls 1.5 m,1.5 m, what is the speed of m2?
A block of mass m1 =3 kg is on a table and attached to a block of mass m2 =4 kg m2 =4 kg by a mass less string. The coefficient of friction is 0.28 between the table and m1. The system is released from rest. After m2 falls 1.5 m,1.5 m, what is the speed of m2?
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter9: Dynamics Of A System Of Particles
Section: Chapter Questions
Problem 9.64P
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Question
A block of mass m1 =3 kg is on a table and attached to a block of mass m2 =4 kg m2 =4 kg by a mass less string. The coefficient of friction is 0.28 between the table and m1. The system is released from rest. After m2 falls 1.5 m,1.5 m, what is the speed of m2?
Expert Solution
Step 1
Draw the free-body diagram of the mass m1.
Step 2
The block of mass m1 doesn’t move along the vertical direction. Thus, the net force along the vertical direction will be zero. Therefore,
The formula for the frictional force acting on the mass m1 is:
Step 3
Apply Newton’s second law of motion along the horizontal direction.
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