Forces and Motion - Ramp Question

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SUMMARY

The discussion focuses on calculating the initial speed required for a mass to ascend a frictionless ramp inclined at 30 degrees with a height of 10 meters. The solution for the first part of the problem yields an initial speed of 14 m/s, while the second part, which includes a static friction coefficient of 0.200, results in an initial speed of 16.2 m/s. The participants emphasize using energy conservation principles, specifically the equation mgh = (1/2)mΔv², as the most effective method for solving these types of problems.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematics equations
  • Knowledge of energy conservation principles
  • Basic trigonometry for resolving forces on inclined planes
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  • Study the application of energy conservation in mechanical systems
  • Learn how to resolve forces on inclined planes using trigonometric functions
  • Explore the effects of friction on motion, specifically static friction coefficients
  • Practice solving problems involving kinematics and dynamics in two dimensions
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of energy conservation and motion on inclined planes.

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Homework Statement


There is a ramp, with an angle of 30 degrees. The height of the ramp is 10 m. There is a mass at the bottom of the ramp. Assuming the ramp is frictionless, determine the initial speed that the mass must have so that it just comes to rest at the top of the ramp.

Part 2 is to repeat with a coefficient of static friction of 0.200.

Homework Equations



I assume all kinematics equations and the use of Newton's law equations.


The Attempt at a Solution



I tried finding the x and y components of velocity by using the height (10m) as distance, v2 = 0m/s and acceleration as 9.8 m/s^2. However, I was not sure how that would work for the horizontal component. I attempted to find the acceleration using Newtons second law; ma = F - mgsin30, however that leaves me with two variables. I was able to get the answer for the first part by using the kinematics equation 2da = v2^2 - v1^2 and using d = 10m, v2 = 0m/s, a = -9.8m/s^2 however I do not know how that works.

The answer for part 1 is 14 m/s, the answer for part 2 is 16.2 m/s.
 
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You can use energy conservation. It's the easiest way.
I don't really understand your attempt.
 
^ It looks like its just mgh=(1/2)mΔv^{2}
 

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