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

[pugola12]

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!

 

[PeterO]

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.

 

[pugola12]

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

 

[PeterO]

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

 

[rwisz]

Your approach and calculations are correct. The mistake may be in the units used. Make sure all units are consistent and in SI units. Also, check for any rounding errors. Another possible issue could be with the angle used in the calculations. Make sure it is in radians instead of degrees. If the angle is given in degrees, you can convert it to radians by multiplying by π/180. I hope this helps.