# Velocity at bottom of ramp

1. Sep 17, 2011

### pugola12

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

Conservation of energy: Ebefore=Eafter

KE=.5mv^2

PE=mgh

3. 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: Sep 17, 2011
2. Sep 17, 2011

### PeterO

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. Sep 17, 2011

### pugola12

4. Sep 18, 2011