The problem involves a 2kg mass sliding down an 8m inclined plane, starting from rest and taking 0.50 seconds to do so. Participants are tasked with calculating the net force acting on the mass along the incline.
The discussion is ongoing, with participants exploring various approaches to understand the problem. Some guidance has been offered regarding the use of kinematic equations, but there is no consensus on the specific steps to take. Participants are actively questioning their understanding and seeking further explanations.
There is mention of limited notes and a lack of diagrams, which may affect participants' ability to visualize the problem. Some participants express frustration with their teacher's support, indicating a challenging learning environment.
Can you work out the acceleration of the object?NickS606 said:I'm sorry I am still new, can I get a better explanation please?
Do you know any Kinematic (SUVAT) equations?NickS606 said:How can I do that?
Really? Then I'm going to go ahead and say take a look in your textbook or class notes and find them.NickS606 said:I'm just going to go ahead and say I have never even heard of those equations.
Perhaps, but we'll have to do some work first.NickS606 said:Ya I don't really recall those. Anyways, would I do [tex]a_{ave} = \Delta v / \Delta t[/tex]
No. Did you just pluck those numbers out of thin air?NickS606 said:I did 12N/3.5s and my answer is 3.4, is this correct?
Okay. Have you looked for the kinematic equations in your notes or text?NickS606 said:I got the numbers out of the question, and there is no diagram, it is height and width for sure though.
Okay, well we cannot use that equation. Try reading http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html" and see if it jogs your memory.NickS606 said:Well there aren't many notes and all I could find was this equation [tex]v_{ave} = \Delta x / \Delta t[/tex]
The highlighted coefficient should be 0.5 rather than 1.5, but yes, this is the equation to use. Can you start filling in the known variables?NickS606 said:I'm just not getting it, I assume the equation to be used is
d=vi*t+1.5*a*t^2
One may use the above formula even if the initial velocity is non-zero.Delphi51 said:Yes, that formula will do it if the initial velocity is zero.
Acceleration is what you want to find out! You need to solve for the acceleration.NickS606 said:Is it d=0*.50s+.5*a*.50s^2
I put a for acceleration because I don't know what acceleration is :/