Using force equations and kinematic equations

[aligass2004]

Homework Statement

A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) 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?

Homework Equations

F = ma
Frictional force = uk(n)
kinematic equations

The Attempt at a Solution

I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).

 

[rootX]

Homework Statement

A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) 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?

Homework Equations

F = ma
Frictional force = uk(n)
kinematic equations

The Attempt at a Solution

I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).

I cannot see friction in your equations(or method), and normal force cannot be 59 N. It is much smaller

 

[aligass2004]

Normal - Wy = m(ay) or Normal - mgcos (theta) = m(ay) right?

 

[rootX]

m(ay) is 0
so
N = mgcos theta

N = 1 * g * cos (t)

and how u got >10?

 

[aligass2004]

The normal force is 7.73N. So now I substitute the normal force into -fricitional force-Wx = m(ax).

 

[learningphysics]

Are you allowed to use work/energy for this problem?

 

[aligass2004]

I don't think so. We haven't gone over work or energy yet.

 

[learningphysics]

The normal force is 7.73N. So now I substitute the normal force into -fricitional force-Wx = m(ax).

yup. this looks right.

 

[aligass2004]

For the acceleration, I got -7.586 m/s^2. I tried using Vxf^2 = Vxi^2 + 2(ax)(delta x), but it didn't work.

 

[lightgrav]

what did you use for the final velocity?
... by the way, this distance is NOT the height!

 

[aligass2004]

Very good point. I hadn't realized that at first. So how do I go about finding the vertical height?

 

[learningphysics]

Very good point. I hadn't realized that at first. So how do I go about finding the vertical height?

use trig. When you get the distance along the ramp... the height is just dsin(38).

 

[aligass2004]

How do I find the distance? By using Vxf^2 = Vxi^2 + 2(ax)(delta x)?

 

[aligass2004]

I'm still confused about this problem.

 

[learningphysics]

How do I find the distance? By using Vxf^2 = Vxi^2 + 2(ax)(delta x)?

Yes. What did you get for distance?

 

[aligass2004]

I got 1.459m for the distance

 

[learningphysics]

I got 1.459m for the distance

How did you get that? Can you show your calculations?

 

[aligass2004]

Well, I assumed that Vxf would be zero since the block would stop on the ramp. Then I used 14 for Vxi and -67.179 for the acceleration.

-196 = 2(-67.179)deltax
-196 = -134.358deltax
delta x = 1.459

 

[learningphysics]

-67 isn't the right acceleration. how did you get that?

 

[aligass2004]

-(ukn)-Wx = m(ax)

 

[aligass2004]

I've got equations everywhere. I think I may have used the wrong acceleration. The acceleration is -7.586. So, is the distance 12.919m?

 

[learningphysics]

I've got equations everywhere. I think I may have used the wrong acceleration. The acceleration is -7.586. So, is the distance 12.919m?

yeah, that looks better. I'm getting acceleration as -[gsin(38) + uk*gcos(38)] = -7.578 m/s^2.

so height is distance*sin(38)

 

[aligass2004]

I got the height to be 7.954 m. So how do I start part b?

 

[learningphysics]

I got the height to be 7.954 m. So how do I start part b?

find the acceleration. then use that to find the velocity.

 

[aligass2004]

How do I find the acceleration?

 

[learningphysics]

How do I find the acceleration?

same way you got the acceleration before.

 

[aligass2004]

Ok, so the forces in the x direction are the frictional force and the x component of the weight. So the equation would be Fr-Wx = m(ax). And I got the answer to be -4.494.

 

[learningphysics]

Ok, so the forces in the x direction are the frictional force and the x component of the weight. So the equation would be Fr-Wx = m(ax). And I got the answer to be -4.494.

looks right. what do you get for the velocity?

 

[aligass2004]

To find the velocity, I used Vxf^2 = Vxi^2 + 2(ax)(delta x). I got -116.116, but now I'm stuck because you can't take the square root of a negative number.

 

[aligass2004]

Nevermind, I got it. The negative sign is just an indication of direction so I got rid of it, took the square root, and tacked it back on for an answer of -10.776.