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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ?
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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string o...
Analysis:
- To determine the minimum weight to stop the motion of the masses, we need to consider the forces acting on the system.
- The forces involved are the tension in the string, the weight of each mass, and the frictional force opposing the motion.
Free Body Diagram:
- For mass m1, the forces acting are tension T in the string and frictional force f.
- For mass m2, the forces acting are tension T in the string, weight mg, and frictional force f.
- The frictional force f can be calculated as f = μ*N, where μ is the coefficient of friction and N is the normal force.
Equations of Motion:
- For mass m1: T - f = m1*a (1)
- For mass m2: T - mg - f = m2*a (2)
Minimum Weight Calculation:
- To stop the motion, acceleration a should be zero.
- From equations (1) and (2), we can calculate the minimum weight m needed to stop the motion.
Conclusion:
- By analyzing the forces, setting up the equations of motion, and calculating the minimum weight required, we can determine the additional weight needed to stop the motion of the masses connected by a string over a frictionless pulley.
- To determine the minimum weight to stop the motion of the masses, we need to consider the forces acting on the system.
- The forces involved are the tension in the string, the weight of each mass, and the frictional force opposing the motion.
Free Body Diagram:
- For mass m1, the forces acting are tension T in the string and frictional force f.
- For mass m2, the forces acting are tension T in the string, weight mg, and frictional force f.
- The frictional force f can be calculated as f = μ*N, where μ is the coefficient of friction and N is the normal force.
Equations of Motion:
- For mass m1: T - f = m1*a (1)
- For mass m2: T - mg - f = m2*a (2)
Minimum Weight Calculation:
- To stop the motion, acceleration a should be zero.
- From equations (1) and (2), we can calculate the minimum weight m needed to stop the motion.
Conclusion:
- By analyzing the forces, setting up the equations of motion, and calculating the minimum weight required, we can determine the additional weight needed to stop the motion of the masses connected by a string over a frictionless pulley.
Community Answer
Two masses m1=5 kg and m2=10 kg, connected by an inextensible string o...
Where is figure??
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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ?
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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ?.
Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ?.
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Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ?, a detailed solution for Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? has been provided alongside types of Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is ? theory, EduRev gives you an
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