What Is the Speed of the Block at the Bottom of the Ramp?

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Homework Help Overview

The problem involves an 11kg block sliding down a frictionless ramp inclined at 30° and then moving along a flat surface. The goal is to find the speed of the block at the bottom of the ramp, using principles of kinematics and conservation of energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the problem using both kinematics and conservation of energy, but expresses confusion over the results being incorrect. They question their logic regarding the calculations for acceleration and height.

Discussion Status

Some participants provide feedback on the calculations, suggesting that the original poster may have misapplied trigonometric relationships in determining acceleration and height. There is an ongoing exploration of the assumptions made in the calculations.

Contextual Notes

Participants note that the problem involves understanding the relationship between gravitational acceleration and the components of forces acting on the block on the ramp. There is a discussion about the implications of using sine functions in the context of the ramp's angle.

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Homework Statement



Starting from rest an 11kg block slides 11.9m down a frictionless ramp (inclined at 30° from the floor) to the bottom. The block then slides an additional 26.3m along the floor before coming to a stop. The acceleration of gravity is 9.8 m/s^2. Find the speed of the block at the bottom of the ramp.


Homework Equations



Vf^2=vi^2+2a(d-do)

Conservation of energy: Ebefore=Eafter

KE=.5mv^2

PE=mgh


The Attempt at a Solution



I tried solving this two different ways, and both gave me the same answer, which was wrong. First I tried kinematics.

Vf^2=vi^2+2a(d-do)
Vf=sqrt(2*9.8/sin30*11.9)

(I used 9.8/sin30 because 9.8 is the vertical component, so 9.8/sin30 is the acceleration down the ramp)

Vf=21.598


The other way I tried used conservation of energy.

h=11.9/sin30=23.8

PE=KE
mgh=.5mv^2
gh=.5v^2
v=sqrt(9.8*23.8/.5)
v=21.598

I don't see where I went wrong in my logic, so I would greatly appreciate some guidance. Thank you!
 
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pugola12 said:

Homework Statement



Starting from rest an 11kg block slides 11.9m down a frictionless ramp (inclined at 30° from the floor) to the bottom. The block then slides an additional 26.3m along the floor before coming to a stop. The acceleration of gravity is 9.8 m/s^2. Find the speed of the block at the bottom of the ramp.


Homework Equations



Vf^2=vi^2+2a(d-do)

Conservation of energy: Ebefore=Eafter

KE=.5mv^2

PE=mgh


The Attempt at a Solution



I tried solving this two different ways, and both gave me the same answer, which was wrong. First I tried kinematics.

Vf^2=vi^2+2a(d-do)
Vf=sqrt(2*9.8/sin30*11.9)

(I used 9.8/sin30 because 9.8 is the vertical component, so 9.8/sin30 is the acceleration down the ramp)

Vf=21.598


The other way I tried used conservation of energy.

h=11.9/sin30=23.8

PE=KE
mgh=.5mv^2
gh=.5v^2
v=sqrt(9.8*23.8/.5)
v=21.598

I don't see where I went wrong in my logic, so I would greatly appreciate some guidance. Thank you!

In the first case you have used 9.8/sin30. It should be 9.8*sin30
In the second case you used h/sin30. That should have been h*sin30.
 
PeterO said:
In the first case you have used 9.8/sin30. It should be 9.8*sin30
Why is that? I used 9.8/sin30 because sin30=9.8/a, so a=9.8/sin30

I see what I did wrong when I found h, but not when I found a.

*sorry I couldn't get the quote to work right
 
pugola12 said:
PeterO said:
In the first case you have used 9.8/sin30. It should be 9.8*sin30
Why is that? I used 9.8/sin30 because sin30=9.8/a, so a=9.8/sin30

I see what I did wrong when I found h, but not when I found a.

*sorry I couldn't get the quote to work right

Apply logic:

The acceleration down a slope is going to be less than the acceleration in free fall. [Indeed, if the slope is very slight, the acceleration is almost zero]

All sine values are fractions [except for 90 degrees when it is 1.

Do you multiply by a fraction or divide by a fraction if you want a smaller answer?

Hopefully you will agree that you multiply to get a smaller answer.

EDIT: when you got sine 30 = 9.8/ a I think you mistakenly assumed 9.8 and a were sides of the triangle that is the ramp The force vector triangle shows sin30 = a/9.8
 

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