Energy Conservation and Friction

In summary: Next, you need to decide what you want h_1 to be. h_1 is the initial height. It can be anything you want. I recommend defining it as the height of the incline above the spring. It is not the height of the incline above the floor, since the spring is at the bottom. Thus h_1=s sin theta.f=[-(0 + 0.5 m v_2^2) + (m g s sin theta + 0)]/ssolving for v_2, I get
  • #1
akhila1489
3
0

Homework Statement


A 2.00 kg is released on a 53.1degree incline, 4.0 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are static friction constant = 0.40, and kinetic friction constant= 0.20. The mass of the spring is negligible.
a)what is the speed of the package just before it reaches the spring.
b)what is the maximum compression of the spring?
c)the package rebounds back up the incline. How close does it get to its initial position?


Homework Equations





The Attempt at a Solution


K1+U1+Wother = K2+U2
Wother= -fs
s is 4m, the distance it needs to get to the spring.

f= fk + fs
fk= 0.20n
fs=0.40n

how do i get n? (n=wcos53.1??)

in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s
f=[-(0+mgh)-(0.5mv^2 + 0)]/4 = 7.06
28.24=(-2*9.8*h)-(0.5*2*v^2)

to solve for v, we need h, is h simple sin53.1?

(*i made a diagram as well, but it isn't allowing me to post it...*)
 
Last edited:
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  • #2
i got part c by doing the following

let x = max compression

U1-fs=0.5kx^2
s= 4+x​
k=120N/m​
2*9.8*(4+x)sin53.1-2.35(4+x)=60x^2
solving the quadratic gives you x=1.06m.

is this remotely correct?
 
  • #3
Welcome to PF!

akhila1489 said:
in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s
f=[-(0+mgh)-(0.5mv^2 + 0)]/4 = 7.06
28.24=(-2*9.8*h)-(0.5*2*v^2)

to solve for v, we need h, is h simple sin53.1?

Hi akhila1489! Welcome to PF! :smile:

Yes, h is 4sin53.1.

but I can't follow the rest of what you've done for part a). :confused:

Just use KE + PE = constant! :smile:
 
  • #4
Hi akhila1489,


akhila1489 said:

Homework Statement


A 2.00 kg is released on a 53.1degree incline, 4.0 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are static friction constant = 0.40, and kinetic friction constant= 0.20. The mass of the spring is negligible.
a)what is the speed of the package just before it reaches the spring.
b)what is the maximum compression of the spring?
c)the package rebounds back up the incline. How close does it get to its initial position?


Homework Equations





The Attempt at a Solution


K1+U1+Wother = K2+U2
Wother= -fs
s is 4m, the distance it needs to get to the spring.

f= fk + fs

You don't want to add the kinetic and static frictional forces. You need to decide which one applies to the case of a object sliding down an incline plane, and use that.

fk= 0.20n
fs=0.40n

how do i get n? (n=wcos53.1??)

in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s

I don't think this follows from your original equation. Your original equation was

[tex]
K_1+U_1+W = K_2+U_2
[/tex]

so this means that K2 and U2 are the final values, and K1 and U1 are the intial values. You also had W= -f s. So when you solve for f, I think you have a sign error; it should be:

[tex]
f=[-(K_2 +U_2) + (K_1 + U_1)]/s
[/tex]

(Also, later you plug in final values for K1 and U1, and initial for K2 and U2.)
 

1. What is energy conservation?

Energy conservation is the principle that states energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

2. How does friction affect energy conservation?

Friction is a force that opposes motion between two surfaces in contact. It converts kinetic energy (energy of motion) into thermal energy (heat), decreasing the total amount of energy in a system. This is why friction is often considered to be an energy-wasting process.

3. Can friction be beneficial for energy conservation?

Yes, friction can be beneficial for energy conservation in certain situations. For example, friction is necessary for objects to stay in place and not slide off surfaces. It can also be used to slow down or stop moving objects, which can be helpful in preventing accidents and saving energy.

4. How can we reduce friction to conserve energy?

There are a few ways to reduce friction and conserve energy. One way is to use lubricants, such as oil or grease, to create a slippery layer between surfaces. Another way is to use smoother materials for surfaces that come into contact with each other. Additionally, keeping surfaces clean and free of debris can also reduce friction.

5. Is energy conservation important for the environment?

Yes, energy conservation is important for the environment because it helps to reduce our overall energy consumption. This means less fossil fuels are burned, leading to less air pollution and greenhouse gas emissions. It also helps to preserve natural resources and protect ecosystems from the negative impacts of energy production.

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