# Kinetic friction of a wood block

1. Apr 20, 2007

### spin360

So the problem states:

A 1.4 kg wood block is launched up a wooden ramp that is inclined at a 22 deg angle. The block's initial speed is 14 m/s. Use Uk = 0.20 for the coefficient of kinetic friction for wood on wood.

U = mu...

(a) What vertical height does the block reach above its starting point?

(b) What speed does it have when it slides back down to its starting point?

Okay so this is how I started it...

I drew a force diagram of the object on an inclined ramp at 22 deg. Set the x axis parallel with the object.. so basically at the same angle. 3 forces.. normal, weight, and kinetic friction.

E = summation
M = mass
A = accel
N = normal
Mg = mass * gravity
fk = kinetic friction

E(Fy) = MAy = 0
N - Mg*cos(22) = 0
N = 12.73

E(Fx) = MAx
-fk - Mg*sin(22) = MAx
-Uk*N - Mg*sin(22) = MAx
-0.20*12.73 - 5.14 = MAx
Ax = -7.686 m/s^2

Vf = Vi + a*t
t = 1.95s

Plug that into the x kinematic equation...
xf = xi + vi*t + (1/2)*a*t^2
= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)
= 12.507 m

The answer is wrong.. if someone could disect my work and figure out what I'm doing wrong that'd be awesome, thanks!

2. Apr 20, 2007

### hage567

You didn't actually divide by the M in the final step to get your acceleration.

Can you elaborate on how you found t? Using your value for a, I don't get the same answer that you did. Maybe you made a typo?

Last edited: Apr 20, 2007
3. Apr 20, 2007

### Staff: Mentor

In addition to what hage567 points out:
Where do the 15 and the sin(22) come from? The initial speed is 14 m/s parallel to the incline. Once you find the distance up the incline using this formula, you'll have to convert it to height.

4. Apr 20, 2007

### hage567

Don't forget that the kinematic equations will give you the distance ALONG the incline, not the height that the block was displaced in the vertical direction.

I'm not understanding the 15*sin(22) part. For one thing, the initial velocity is 14 m/s.

EDIT: I'm too slow.

5. Apr 20, 2007

### spin360

Ahh good catch. Okay then I get -5.49 m/s^2 for the accel and 2.73 for t. I plugged that into the y formula and get -5.12 which is obviously not right

6. Apr 20, 2007

### spin360

Okay... so then would just be y = 14*2.73 + (1/2)(-5.49)(2.73^2)?

Actually yeah that's wrong too

7. Apr 20, 2007

### spin360

I got 17.76 m for the "distance" not sure how to convert to height?

8. Apr 20, 2007

### spin360

haha nevermind i got it... 17.76*sin(22) = 6.65m for the height thanks guys!