Block sliding down incline shaped as circle

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SUMMARY

The discussion focuses on the physics problem involving a 2-kilogram block sliding down a frictionless quarter-circle incline and then onto a horizontal plane. The block experiences an 8 Newton frictional force while sliding horizontally, resulting in an acceleration of 4 m/s². The time taken for the block to slide horizontally is calculated to be 2 seconds using the correct SUVAT equation: s = vt + 0.5at². The radius of the incline is yet to be determined, but the problem-solving approach emphasizes the importance of correctly applying kinematic equations.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of kinematic equations (SUVAT equations)
  • Basic principles of friction and motion
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation and application of SUVAT equations in various motion scenarios
  • Explore the effects of friction on motion and how to calculate net forces
  • Learn about circular motion and its relationship to inclined planes
  • Investigate the concept of energy conservation in frictionless systems
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Students studying physics, particularly those focusing on mechanics and kinematics, as well as educators looking for examples of problem-solving in motion dynamics.

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Homework Statement


A 2-kilogram block is released from rest at the top of a curved incline in the shape of a quarter circle of radius R. The block then slides onto a horizontal plane where it finally comes to rest 8 meters from the beginning of the plane. The curved incline is frictionless, but there is an 8 Newton force of friction on the block while it slides horizontally. Assume g=10N/kg
A) Determine the magnitude of the acceleration of the block while it slides along the horizontal plane.
B) what time elapses while the block is sliding horizontally
C)Calculate the radius of the incline in meters




Homework Equations


F/m=a
t= 2(delta)x/a (squared)



The Attempt at a Solution


a- 8N/2kg=4m/s^2
b- T=2(8m)/4m/s^2 (squared)= 2 sec
c-?
 
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a) OK

b) I don't know what equation you are using there, but it's wrong and doesn't give the answer of 2 seconds. 2 seconds is the right answer (as per my equation below), but your equation gives an answer of 1 second or 16 seconds depending on how you did it (see example below).

t = 2(8)/42 = 16/16 = 1
or you squared everything
t = (2(8)/4)2 = (16/4)2 = 16
Both of these are wrong. It appears you have rearranged the below equation incorrectly.

You want to use the SUVAT equations of motion. In this case: s=vt+0.5at2, where s = distance traveled (8m), v = final speed (0m/s) and a = acceleration (from part a).

Substitute in the values and you end up with s=0.5at2. Rearrange and that will give you the travel time.

Based on my SUVAT equation above, yours should be: Sqrt(2s/a) = t.
 

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