Solving Power Energy Problem: Inclined Plane 30 Degrees, 6.4m Long

  • Thread starter Thread starter deviantdevil
  • Start date Start date
  • Tags Tags
    Energy Power
Click For Summary
SUMMARY

The discussion focuses on solving the problem of a book sliding down a 6.4m inclined plane at a 30-degree angle, with a kinetic coefficient of friction of 0.2. The initial approach utilized a combination of energy and Newtonian mechanics, specifically the equation 6.4mg(sin(30))=0.5mv^2+0.2mg(cos(30))6.4. Participants clarified that this equation represents the balance of initial potential energy, final kinetic energy, and work done against friction. The conversation emphasizes the need for a purely energy-based solution to the problem.

PREREQUISITES
  • Understanding of potential and kinetic energy concepts
  • Familiarity with friction coefficients and their impact on motion
  • Knowledge of basic trigonometry, specifically sine and cosine functions
  • Ability to apply energy conservation principles in physics
NEXT STEPS
  • Research energy conservation principles in mechanics
  • Study the effects of friction on inclined planes in physics
  • Learn how to derive equations of motion using energy methods
  • Explore advanced problems involving inclined planes and energy considerations
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in applying energy conservation principles to solve motion problems involving friction.

deviantdevil
Messages
3
Reaction score
0
Q: An inclined plane at 30 degrees is 6.4m long. A book, which has a kinetic coefficient of friction with the incline of 0.2, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline?

my solution is as follows:

6.4mg(sin(30))=0.5mv^2+0.2mg(cos(30))6.4
cancel out m and find v.
use vf^2=vi^2+2ad to find a
use d=vit+0.5at^2 to find t.

i think this solution works, but as you can see, it doesn't really "use energy considerations"...
any idea how i should go about solving this problem?
 
Physics news on Phys.org
deviantdevil said:
i think this solution works, but as you can see, it doesn't really "use energy considerations"...

What did you use to get this line: "6.4mg(sin(30))=0.5mv^2+0.2mg(cos(30))6.4"

It looks to me like: initial potential energy = final kinetic energy + work done by friction
Why do you not consider that using energy arguments?
 
i assumed the question to be asking for a solution that only requires energy arguments. as you can see, i go into Newtonian mechanics in the latter part of the solution. Is there a way to solve this question using energy/power alone?
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 7 ·
Replies
7
Views
7K