Deriving Speed at an Inclined Plane

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Homework Help Overview

The discussion revolves around a block sliding down a frictionless inclined plane with a specific angle and length. The original poster attempts to determine the speed of the block at the bottom of the incline using kinematic equations but encounters difficulties in their calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of energy conservation versus kinematics to find the speed of the block. Some suggest that energy conservation is more appropriate, while others reflect on the challenges faced with kinematic equations. There are questions about the role of mass in the equations and whether kinematics can be effectively applied in this scenario.

Discussion Status

There is an ongoing exploration of different methods to solve the problem, with some participants providing guidance on using energy conservation. However, there is no explicit consensus on the best approach, and multiple interpretations of the problem are being considered.

Contextual Notes

Participants note that the mass of the block is not given, which raises questions about its relevance in the equations discussed. There is also mention of the potential difficulty in applying kinematic equations effectively.

zcabral
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Homework Statement


A block slides down a frictionless plane having an inclination of = 12.2° (Fig P5.22). The block starts from rest at the top and the length of the incline is 2.85 m.


Homework Equations


F=ma
xf=xi+vit+1/2at^2



The Attempt at a Solution


i tried to solve this by using the kinematic equation above. i know the acceleration is 2.07 m/s^2 so i plugged it into find t in order to find speed but it didnt work. I am totally stuck. i know if i find t i could find the answer by absolute value v=dx/dt
 
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If you're searching for the velocity of the block at the very bottom of the ramp, you actually utilize Energy conservation instead of kinematics.

PE = mgh
KE = 1/2*mv^2

The block is at rest on top of the incline plane, which means no KE and only PE. That means it starts out with a total E of m*9.8*(2.85*sin 12.2)

At the bottom of the ramp, all that PE has been converted into KE.

KE = 1/2(m)(v^2)

So since E is conserved because there is no friction on the ramp nor air drag, set PE = KE and solve for V. Note that your masses will cancel out.
 
the mass is not given
 
zcabral said:
the mass is not given

no problem, if the mass is the same in the begin and in the end, they cut in the equation... state the equation on a paper and you will see
 
although its much easier to do by consrevation of energy, but you should also reflect on why kinematics did not work for you. Are you still weak in kineatics? You may have escaped from using kinematics from this question but not all questions can be solved by conservation of energy,The correct equation to use would be v^2=u^2+2as. if you used the equation in your OP, then you should have solved quadratically for t and substituted t into (v-u)=at.
 
this should be enough...

velocity at the bottom of an inclined plane is independent of angle of inclination...

infact it is given by...

v=[(2)gh]^1/2
 
Last edited:
physixguru said:
this should be enough...


v=[(2)gh]^1/2


I think it would be way better to derive this result rather than to mug this equation and plug the numbers into it
 
Oerg said:
I think it would be way better to derive this result rather than to mug this equation and plug the numbers into it

well you are right but is it a too difficult equation my frnd?
 

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