Calculating Acceleration with Friction and Applied Force

  • Thread starter Thread starter dance_sg
  • Start date Start date
  • Tags Tags
    Force
Click For Summary
SUMMARY

The discussion focuses on calculating the acceleration of a sled and cargo weighing 490 N, towed by a tractor exerting a force of 300 N, with a coefficient of sliding friction of 0.400. The user correctly converts the weight from Newtons to kilograms by dividing by 9.81, resulting in a mass of approximately 49.95 kg. The frictional force is calculated using the formula Fk = μkN, leading to the final acceleration calculation using the formula Acceleration = (F - fk)/m. The solution confirms the correct application of physics principles in determining the sled's acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of friction coefficients and their application
  • Ability to convert weight from Newtons to kilograms
  • Familiarity with basic algebraic manipulation for solving equations
NEXT STEPS
  • Study the effects of varying coefficients of friction on acceleration
  • Learn about the dynamics of forces in horizontal motion
  • Explore advanced applications of Newton's second law in real-world scenarios
  • Investigate the impact of mass on acceleration in different contexts
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and friction in practical applications.

dance_sg
Messages
113
Reaction score
0

Homework Statement



A sled and cargo weighing 490 N is towed by a tractor over a horizontal field where the coefficient of sliding friction is 0.400. If the tractor exerts a force of 300 N, what is the acceleration of the sled?

Homework Equations


Fk/µkM


The Attempt at a Solution


i Tried changing the 490N to Kg by dividing it by 9.81. that gave me 49.95. i then multiplied that by 0.400. then i divided that final number from 300 N.
 
Physics news on Phys.org
dance_sg said:

Homework Statement



A sled and cargo weighing 490 N is towed by a tractor over a horizontal field where the coefficient of sliding friction is 0.400. If the tractor exerts a force of 300 N, what is the acceleration of the sled?

Homework Equations


Fk/µkM


The Attempt at a Solution


i Tried changing the 490N to Kg by dividing it by 9.81. that gave me 49.95. i then multiplied that by 0.400. then i divided that final number from 300 N.

490 N is the normal reaction. So the frictional force fk= μkN.
Acceleration = (F - fk)/m
 
O thank you so much :)
 

Similar threads

Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K