Kinetic friction of a wood block

[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!

 

[hage567]

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!

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?

 

[Doc Al]

In addition to what hage567 points out:

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

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.

 

[hage567]

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

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.

 

[spin360]

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?

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

 

[spin360]

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

Actually yeah that's wrong too

 

[spin360]

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

 

[spin360]

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