Finding Tension and Power in an Eskimo's Sled Pull

  • Thread starter Thread starter dman_PL
  • Start date Start date
  • Tags Tags
    Power Tension
Click For Summary
SUMMARY

The discussion focuses on calculating the tension in a rope and the power due to friction when an Eskimo pulls a 100kg sled at a constant speed of 1.5m/s with a force applied at a 30° angle. The correct tension in the rope is determined to be 113N, derived from the horizontal component of the tension needed to overcome a frictional force of 98.1N. The power dissipated due to friction at this speed is calculated to be 147W. The initial approach to the problem was incorrect, highlighting the importance of accurately applying physics equations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of kinetic friction and its coefficient
  • Familiarity with basic trigonometry, specifically cosine functions
  • Ability to calculate power using the formula Power = Force x Velocity
NEXT STEPS
  • Study the principles of tension in ropes and angles in physics problems
  • Learn about the effects of friction on motion and energy dissipation
  • Explore the application of trigonometric functions in physics calculations
  • Investigate the relationship between work, power, and energy in mechanical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of tension and power calculations in real-world scenarios.

dman_PL
Messages
15
Reaction score
0

Homework Statement



An Eskimo pulls a 100kg sled with a constant force through the snow
at a constant speed of 1.5m/s. He applies this force by pulling on a rope at 30◦
to the horizontal. There is a coefficient of kinetic friction between the sled and the snow of 0.1

1)Find the tension in the rope
2)What is the power due to the frictional force (rate at which energy is being dissipated)
the instant he stops pulling on the sled

Homework Equations


The Force of Tension = (mass)(acceleration) - (Coefficient of Friction)(-mass · gravity)
2) Power= Work/Time


The Attempt at a Solution


1) I thought it was a simple just plug it in, so i did Fτ=(100kg)(1.5m/s)-(.1)(-100kg*9.8)=248N
2) I used the 248N to find the amount of work that was done, so Work=248N*1.5m/s*Cos(30)=322.16. However after this I am confused as to how i find time.
 
Physics news on Phys.org
A sled of MASS 100kg has a WEIGHT of 100 x 9.81N =981N
A coeff of friction of 0.1 means that the force of friction between the sled and the snow is 0.1 x 981N = 98.1N
So to pull the sled at (ANY) CONSTANT SPEED (velocity) needs a horizontal force of 98.1N
This force is provided by the horizontal component of the tension in the rope
So T x Cos30 = 98.1 gives T = 98.1/Cos30 = 113N
Power = force x velocity
So with a friction force of 98.1N at a velocity of 1.5m/s the power is 98.1 x 1.5 = 147W
 
Thanks :) i guess my initial equation was off
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
Replies
3
Views
2K
Replies
1
Views
5K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 15 ·
Replies
15
Views
2K