# Length of Ramp: 12.7m

• neongoats

## Homework Statement

You are passing a construction site on the way to physics class, and stop to watch for a while. The construction workers appear to be going on a coffee break, and have left a large concrete block resting at the top of a wooden ramp connecting one of the building's upper floors to the ground. As soon as their backs are turned, the block begins to slide and takes about 2 seconds to reach the ground. You wonder how long the ramp is. You estimate the ramp is at an angle of about 37* to the horizontal, and your physics book gives the coefficient of kinetic friction (uk) between concrete and wood as 0.25

a= F/m
vf=vi+at
vf^2=vi^2+2ax

## The Attempt at a Solution

I found that:
Fnety = Fn - Fgy = Fn - mgsin(37) = 0
Fn = mgsin(37)
Fnetx = Ff - Fgx = uk(mg)sin(37) - mgcos(37) = ma
a = uk(g)sin(37) - gcos(37)
a = .25(-9.8)sin(37) + 9.8cos(37)
a = 6.35 m/s^2

vf-vi = at
vf = 6.35(2)
vf = 12.7 m/s

vf^2 - vi^2 = 2ax
12.7^2 = 2(6.35)x
x = 12.7m

I don't understand how this is wrong?

Did you draw a free-body diagram?
Check your trig relations - the angle is measured from the horizontal.
Don't forget to define the x and y directions as part of your working.

Did you draw a free-body diagram?
Check your trig relations - the angle is measured from the horizontal.
Don't forget to define the x and y directions as part of your working.

I did draw a free body diagram and I believe my trig relations are correct.
After further digging I found nearly the exact same question http://courseweb.stthomas.edu/physics/academics/111/Brita/Spring%202012/Unit%202/GP6%20P11%20S12.pdf but when I calculate the velocity with their numbers I still get the wrong velocity?
Edit: As in their paper says vf = 2.5 m/s but I get 80.36 when I plug those exact same numbers into my calculator. Did they do it wrong or am I messing up my calculation somewhere?

a= .25(-9.8)sin(37) + 9.8cos(37)
I think you have mixed up your sines and cosines. You need a large, clear diagram to work on, or you probably will get these wrong.

Is the correct answer around 7.5 roughly?

I did draw a free body diagram and I believe my trig relations are correct.
Looking at your FBD: if you have theta as the angle of the ramp to the horizontal, and you have defined the +y axis to be upwards and perpendicular to the ramp, then the normal force to the ramp is ##\vec N = \hat\jmath mg\cos\theta##. You may need to redraw your diagram to see this.