A block sliding down an incline

Click For Summary
SUMMARY

The discussion focuses on the physics of a block sliding down an incline and subsequently on a flat surface. The block starts from a height of 4.7 meters on a 30-degree incline, resulting in a speed of 6.887937 m/s at the bottom. The coefficient of friction is 0.28, which affects the block's deceleration on the ground. Participants suggest using kinematic equations and work-energy principles to calculate the distance the block will slide before coming to rest.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = v0 + at
  • Knowledge of forces, including friction and gravitational components
  • Familiarity with work-energy principles in physics
  • Basic proficiency in solving equations involving acceleration and distance
NEXT STEPS
  • Calculate the deceleration due to friction using f = μN
  • Learn how to apply the work-energy theorem to solve for distance
  • Explore kinematic equations, specifically v² = u² + 2as, for motion analysis
  • Investigate the effects of varying coefficients of friction on sliding distance
USEFUL FOR

Physics students, educators, and anyone interested in understanding dynamics and motion on inclined planes and flat surfaces.

Momentum09
Messages
69
Reaction score
0
A block starts from rest at a height of 4.7m on a fixed inclined plane tilted at 30 degrees. The coefficient of friction is 0.28. If the block continues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground until coming to rest?


2. V = V0+ at



3. I found out the speed of the block at the bottom of the ramp, which is equal to 6.887937 m/s. I led Vf = 0, V0 = 6.887937, and a =2.5236 [solved from gsin(delta) - ugcos(delta). Solve for t, and plug this value into the delta x = volt + 1/2 at^2 equation. I am not sure if I'm doing the right thing.

Thank you so much!
 
Physics news on Phys.org
How did you get to that speed at the bottom of the ramp? Show your calculations please.
 
3. I found out the speed of the block at the bottom of the ramp, which is equal to 6.887937 m/s. I led Vf = 0, V0 = 6.887937, and a =2.5236 [solved from gsin(delta) - ugcos(delta). Solve for t, and plug this value into the delta x = volt + 1/2 at^2 equation. I am not sure if I'm doing the right thing.

Thank you so much!


Either calculate the force due to friction and hence the deceleration and use [tex]\ \ v^2=u^2+2as[/tex] or use the work energ concept, [tex]\ \ W_{nc}=E_f-E_i[/tex], where W_nc is the work due to friction and [tex]\ \E_f-E_i[/tex] will be same as [tex]K_f-K_i[/tex]
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 27 ·
Replies
27
Views
10K
Replies
12
Views
2K
Replies
4
Views
4K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 45 ·
2
Replies
45
Views
8K
Replies
6
Views
2K
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K