Solving Block on Incline Physics Problem

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SUMMARY

The discussion focuses on solving a physics problem involving a block on an incline. Participants analyze the free body diagram (FBD) and the equations of motion, specifically addressing the forces acting on the block, including gravitational force (Fg), kinetic friction (fk), and normal force (Fn). Key equations discussed include Fnet = -ma and the correct resolution of forces into their x and y components. The consensus is that the user must correct the orientation of gravitational components to achieve accurate results.

PREREQUISITES
  • Understanding of free body diagrams (FBD)
  • Knowledge of Newton's second law (Fnet = -ma)
  • Familiarity with kinetic friction and normal force concepts
  • Ability to resolve forces into components (x and y)
NEXT STEPS
  • Review the principles of free body diagrams in physics
  • Study the application of Newton's laws in inclined plane problems
  • Learn about the calculations of kinetic friction and its effects on motion
  • Practice resolving forces into components with various incline angles
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators seeking to clarify concepts related to forces and motion.

Maiia
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Can someone tell me if I'm doing this problem right?

Homework Statement


Here is the picture of the problem:
http://i210.photobucket.com/albums/bb23/Kyashii628/block.jpg

In my FBD, I have Fg pointing down, Fgy perpendicular to the incline and pointing down, Fgx parallel to incline pointing down the ramp (to left), kinetic friction pointing up the ramp (to right), and normal force pointing opposite from Fgy.

Homework Equations


Fnet= -ma
fk- Fgx= -ma
u(Fn)= -ma
u(345.0615986)= 648.9664808 -240
u= 1.182397583
 
Last edited by a moderator:
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Maiia said:
Can someone tell me if I'm doing this problem right?

Homework Statement


Here is the picture of the problem:
http://i210.photobucket.com/albums/bb23/Kyashii628/block.jpg

In my FBD, I have Fg pointing down, Fgy perpendicular to the incline and pointing down, Fgx parallel to incline pointing down the ramp (to left), kinetic friction pointing up the ramp (to right), and normal force pointing opposite from Fgy.
good, but remember in your FBD, that once you resolve Fg into its x and y components, you should 'cross out' the Fg downward force.

Homework Equations


Fnet= -ma
fk- Fgx= -ma
good
u(Fn)= -ma
I'm sure you meant u(Fn) = fk
u(345.0615986)= 648.9664808 -240
you have your mg(sin theta)and mg(cos theta) reversed.
u= 1.182397583
Correct your error and your result should be fine.
 
Last edited by a moderator:

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