Solving Physics Problems: Tension, Friction, and Acceleration Explained

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SUMMARY

This discussion focuses on solving a physics problem involving two blocks connected by a string on a frictionless surface, with gravity set at 9.8 m/s². The key equations utilized include Newton's second law, F=ma, to determine the tension in the string and the acceleration of the blocks. It is established that the tension will be less than the applied force of 50.8 N due to the system's dynamics. Additionally, when friction is introduced with a coefficient of 0.0958, the new acceleration and tension must be recalculated, reflecting the impact of friction on the system.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of tension in strings and forces in systems
  • Familiarity with friction coefficients and their effects on motion
  • Basic problem-solving skills in physics
NEXT STEPS
  • Calculate tension in systems with multiple forces acting on them
  • Learn about the effects of kinetic friction on acceleration
  • Explore the concept of systems of particles in physics
  • Practice solving similar physics problems involving tension and friction
USEFUL FOR

Students studying physics, particularly those tackling mechanics problems involving tension, friction, and acceleration. This discussion is beneficial for anyone seeking to enhance their understanding of force interactions in multi-body systems.

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Physics Problem. Please HELP!

Homework Statement


Two blocks on a frictionless horizontal surface are connected by a light strong. Gravity is 9.8.
Diagram:
4.13kg--T--21.2kg--->50.8N
a. What is the tension in the string between the blocks? Answer in units of N.
b. If the surface were frictional, and the coefficient of kinetic friction between each block and the surface is 0.0958, what would be the new acceleration? Answer in units of m/s^2

c. What would be the new tension in the string between the blocks? Answer in units of N




Homework Equations




The Attempt at a Solution


I thought that because it was frictionless for a. the tension would just be 50.8 again, but it was wrong...(im entering answers online)

grrr...I'm completely lost!

Help!? anyone?
 
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No, the tension between the two masses will be less. Just remember that F=ma, and both masses have the same acceleration. So there's a combination tension in the right string, and a single tension in the left string. Makes sense now?
 
Right; if the tension was somehow the same, then the force would be equal on both sides of the rightmost block and it would not move. Consider both masses as a single system first, then look at them separately. It took me a little while to get the hang of force calculations but once I really understood them it's easy as pie.
 

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