What Is the Spring Constant and Velocity of a Box on an Inclined Plane?

[tachu101]

Homework Statement

Given the incline plane, see picture, a box is let loose, attached to a spring in a relaxed position, the box moves down 2.75 meters. The variables: incline is at 32 degrees, the friction coefficient is .125 and the mass of the box is 17.4 kg. The length it slides is 2.75 meters.

Find the: spring constant (k) when friction is present?

spring constant (k) when friction is not present?

the velocity of the box when it is half way down the incline (so 1.375 meters)?

Homework Equations

The teacher said that I could use conservation of energy to solve this and said that I could set it up in the following way

Work(in)= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-Vo^2)+(friction)(distance)+(1/2)(k)(change in spring position^2)

I think that the Work goes to 0 , the Initial Velocity goes to 0, and the initial spring position is 0 ---- so----

0= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-0)+(friction)(distance)+(1/2)(k)(change in spring position^2)

Is there a less cluttered way of solving this problem?

The Attempt at a Solution

The initial height should be (2.75)(sin32)? Right?
The (distance) for the friction, and spring should be 2.75 because that is how far it goes?
Does the final velocity also go to 0, I am not sure about that?

 

[tachu101]

Does anyone have an idea on how to go about solving this question?

 

[Doc Al]

Perhaps it will be easier to apply conservation of energy if you think of it like this:
(initial mechanical energy) + (work done by friction) = (final mechanical energy)

Mechanical energy is the sum of kinetic energy, gravitational PE, and spring PE. The work done by friction (where it exists) will be negative.

To find the work done by friction, first find the friction force.

 

[tachu101]

Can anyone else get the same answers as me ---

k with friction -- 92.1179
velocity half way -- 3.587 m/s
k without friction -- 105.277

 

[tachu101]

Anyone?

 

[Doc Al]

Why don't you show exactly how you got your answers.

 

[tachu101]

I have gotten to

mgh=fs+1/2kx^2 ---- so ---- (17.4)(9.81)(2.75cos32)=(.125(17.4)(9.81)cos32)(2.75)+(1/2)(k)(2.75)^2
SO k= 92.117

then

I replug it into
Work(in)= (mass)(gravity)(height final -height initial)+(1/2)(mass)(vf^2-Vo^2)+(friction)(distance)+(1/2)(k)(change in spring position^2)
To get a velocity of 3.587 m/s

Then to get without friction I just take friction out of the equation to get
mgh=1/2kx^2 where k then equals 105.277

 

[Doc Al]

I have gotten to

mgh=fs+1/2kx^2 ---- so ---- (17.4)(9.81)(2.75cos32)=(.125(17.4)(9.81)cos32)(2.75)+(1/2)(k)(2.75)^2
SO k= 92.117

The change in height should be 2.75*sin32, not cos32.

 

[tachu101]

thank you I don't know what i was thinking there