Need an explanation on how to do a question applying Newtons Laws

Click For Summary
SUMMARY

This discussion focuses on applying Newton's Laws to a problem involving two masses connected by a string over a frictionless pulley. The key calculations involve determining the tension in the string and the acceleration of the 2.0 kg mass on the table. The correct answers are a tension of 14 N and an acceleration of 7.0 m/s². The forum emphasizes the importance of analyzing each mass separately and recognizing that the acceleration is the same for both masses.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with the concept of tension in strings
  • Knowledge of basic physics equations, specifically ΣF=ma
  • Ability to analyze forces acting on objects in a system
NEXT STEPS
  • Study the derivation of tension in pulley systems
  • Learn about free-body diagrams for analyzing forces
  • Explore advanced applications of Newton's Laws in multi-body systems
  • Investigate the effects of friction in pulley systems
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of connected masses in motion.

Asimans
Messages
1
Reaction score
0

Homework Statement


A 2.0 kg mass , placed on a smooth, level table, is attached by a light string passing over a frictionless pulley to a 5.0 kg mass hanging freely over the edge of a table. Calculate...
a) the tension in the string
b) the acceleration of the 2.0 kg mass on the table.The answer given was
T=14N
a=7.0m/s^2

Homework Equations


ΣF=ma

The Attempt at a Solution


ΣF=ma
Ft + Fg = (5+2)a
Ft + (mg) = (7)9.8
Ft + (49) = (68.6)
Ft = 19.6N
 
Physics news on Phys.org
Asimans said:

Homework Statement


A 2.0 kg mass , placed on a smooth, level table, is attached by a light string passing over a frictionless pulley to a 5.0 kg mass hanging freely over the edge of a table. Calculate...
a) the tension in the string
b) the acceleration of the 2.0 kg mass on the table.


Homework Equations


ΣF=ma


The Attempt at a Solution


ΣF=ma
Ft + Fg = (5+2)a
Ft + (mg) = (7)9.8
Ft + (49) = (68.6)
Ft = 19.6N
Hello, Asimans, welcome to PF! Don't try to do this all at once...look at each block separately. The acceleration is not 'g'. Note that the tension in a cord wrapped around an ideal pulley is the same on both sides of the pulley. Also note that the magnitude of the acceleration of each block must be the same in this example. So look at the lower block first, identify the forces acting on it, and apply your relevant equation. Then look at the other block, and do the same. What do you get?
 

Similar threads

Replies
13
Views
3K
Replies
44
Views
4K
  • · Replies 21 ·
Replies
21
Views
11K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K