What is the final velocity of an object sliding down a frictionless ramp?

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Discussion Overview

The discussion revolves around calculating the final velocity of an object sliding down a frictionless ramp, specifically focusing on the application of energy methods and linear motion methods. Participants explore the relevant formulas and engage in problem-solving related to gravitational potential energy and kinetic energy.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving a 2250kg object sliding down a 10m frictionless ramp at a 10-degree angle, seeking to find its final velocity.
  • Another participant asks for the formulas for gravitational potential energy and work, leading to the mention of PE = mgh and W = Fd.
  • There is a correction regarding the formulas, with participants clarifying that kinetic energy is given by 1/2 mv².
  • Confusion arises regarding the calculation process, with one participant mistakenly combining terms in the kinetic energy equation.
  • Participants discuss the importance of maintaining clarity in notation, suggesting that using (1/2)m instead of 1/2m can prevent misinterpretation.
  • One participant emphasizes the need to perform algebraic manipulations before substituting numerical values to avoid errors.
  • There is a suggestion to always include units in calculations to reduce confusion and mistakes.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the application of energy principles and the correct manipulation of formulas. There is no consensus on the correct approach to solving the problem, and confusion persists about the calculations and notation used.

Contextual Notes

Participants highlight the importance of correctly applying formulas and maintaining clarity in calculations. There are unresolved issues regarding the interpretation of terms and the proper sequence of operations in solving the problem.

Taidhg
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I have a question which asks me if an object with a mass of 2250kg starts from rest at the top of a frictionless ramp 10m long and is at an angle of 10 degrees horizontally, what is the velocity of the object the moment it reaches the bottom of the ramp?
I have been trying to work out this question for hours but it baffles me. It does say It can be calculated using the energy method or the linear motion method but I don't properly understand them.
 
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What is the formula for gravitational potential energy and the formula for work?
 
Dale said:
What is the formula for gravitational potential energy and the formula for work?
PE = mgh and W = Fd
 
Oops. I meant the formula for gravitational PE and kinetic energy. Sorry.
 
Dale said:
Oops. I meant the formula for gravitational PE and kinetic energy. Sorry.
Kinetic energy is 1/2 m v2 but I don't have velocity
 
Taidhg said:
Kinetic energy is 1/2 m v2 but I don't have velocity
Right, it is what you want to calculate
 
Dale said:
Right, it is what you want to calculate
So because GPE = 220725
v2 = 220725-1/2m?
 
Dale said:
Right, it is what you want to calculate
my calculations were 220725-½2250 = 219600squared
root 219600 = 468.6 m/s2
I feel like I am doing something wrong.
 
Taidhg said:
v2 = 220725-1/2m?
Did you really mean to write 220725 minus 1/2m?

Kinetic energy is 1/2m times v2, not 1/2m plus v2.

By the way, most people here will read 1/2m as "one divided by 2m", not "one-half m". Better to write it as (1/2)m or more simply m/2.
 
  • #10
jtbell said:
Did you really mean to write 220725 minus 1/2m?

Kinetic energy is 1/2m times v2, not 1/2m plus v2.

By the way, most people here will read 1/2m as "one divided by 2m", not "one-half m". Better to write it as (1/2)m or more simply m/2.
I see what I have done, I got the right answer at last. Thanks for the help :)
 
  • #11
Also, it's a good idea to "first do the algebra, then do the arithmetic." In this case, start with (1/2)mv2 = mgh, then rearrange the equation to put v all by itself on the left, and finally plug in the numbers and calculate the result in one go on your calculator. See what happens to m?
 
  • #12
Taidhg said:
So because GPE = 220725
v2 = 220725-1/2m?
Where are you getting the ##-\frac 1 2 m##? Check your units. Does that make sense? Can you subtract mass from energy?

Always include your units, what you have written is confusing and including units will reduce the confusion and avoid mistakes like this.
 
  • #13
Taidhg said:
my calculations were 220725-½2250 = 219600squared
root 219600 = 468.6 m/s2
I feel like I am doing something wrong.
Stick with the algebraic representation and re-arrangement until the very end and then substitute the numbers that you have. That way you will be able to solve any such question.
 

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