Velocity at bottom of ramp

  • Thread starter pugola12
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  • #1
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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 differant 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!
 
Last edited:

Answers and Replies

  • #2
PeterO
Homework Helper
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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 differant 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!

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


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.
 
  • #3
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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
 
  • #4
PeterO
Homework Helper
2,426
48
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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