Conservation of Energy, Down an Incline with a Spring

In summary: The answer is that the block would reach its maximum speed at the halfway point between the point of first contact and the point where the block's speed is greatest.
  • #1
RavenBlackwolf
15
0

Homework Statement


A 4.0 kg block starts at rest and slides a distance d down a frictionless 35.0
char0E.png
incline, where it runs into a spring. The block slides an additional 16.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 429 N[PLAIN]https://homework2.math.pitt.edu/adm/jsMath/fonts/cmmi10/alpha/100/char3D.pngm .
upload_2017-2-21_5-4-3.png

a) What is the value of d?
b) What is the distance between the point of first contact and the point where the block's speed is greatest?

Homework Equations


Ui+Ki=Uf+Kf
US=1/2kx2
UG=mgh
K=1/2mv2

The Attempt at a Solution


a) This one I got
1/2kx^2=mgΔh
1/2(429)(.162)=mgΔh
Δh=.244
d=.244-.16=.084m (correct answer)
b) This one I'm not sure of
1/2mv2+mgh=1/2mvf2+mg((.16-x)sin(35))+1/2kx2
I got that the velocity at contact is .972 m/s but how do I get the final velocity value? I need it to use the conservation of energy law the way I set it up. I've found similar questions to this online but none of them provide actual explanations/calculations for this part. I'm aware that it accelerates still once it hits the spring but then what?
 

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  • #2
RavenBlackwolf said:
I'm aware that it accelerates still once it hits the spring but then what?
That's the key. As long as it's accelerating (downward), it's speed continues to increase. So what condition must exist at the point of maximum speed?
 
  • #3
Doc Al said:
That's the key. As long as it's accelerating (downward), it's speed continues to increase. So what condition must exist at the point of maximum speed?
The acceleration in the x direction would be 0 correct? I thought of that before but I got lost because wouldn't I need time if I'm using the kinematics equations? Are they what I should use to find vf or should I be using an energy concept?
 
  • #4
RavenBlackwolf said:
The acceleration in the x direction would be 0 correct?
Right. Now figure out where (not when) the acceleration would be zero. Then use energy methods to find the speed.
 
  • #5
Doc Al said:
Right. Now figure out where (not when) the acceleration would be zero.
Where as in Δx? Don't I need the vf to find that though? That's the only reason I wanted to find vf at all. All I have is the .16m when the spring is compressed to the fullest and v=0 I believe ax is zero there too but that can't be the answer. I feel like I'm missing something but I can't figure out what because I've been working this problem too long. Its x acceleration is positive (at least the way I'm modeling it) when it hits the spring then it gets smaller, hits zero, then goes negative. I don't understand where I'm gathering distances from this though. The value I need is the distance at which the acceleration is zero and the velocity is maximized. Would it be halfway down then?
 
  • #6
RavenBlackwolf said:
Don't I need the vf to find that though? That's the only reason I wanted to find vf at all.
You have no need to actually find the max speed, just the position where it is attained.

RavenBlackwolf said:
All I have is the .16m when the spring is compressed to the fullest and v=0 I believe ax is zero there too but that can't be the answer.
If the acceleration were zero at that point, it would just sit there. But it doesn't.

RavenBlackwolf said:
The value I need is the distance at which the acceleration is zero and the velocity is maximized. Would it be halfway down then?
Don't guess. Hint: Use dynamics. Analyze the forces as the spring is compressed.
 
  • #7
Doc Al said:
Then use energy methods to find the speed.
You can skip this step, since you're not asked to find the max speed. :smile:
 

FAQ: Conservation of Energy, Down an Incline with a Spring

How does the conservation of energy apply to an object moving down an incline with a spring?

The conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this scenario, the potential energy of the object at the top of the incline is converted into kinetic energy as it moves down the incline. The spring then absorbs some of this kinetic energy, storing it as potential energy in the form of elastic potential energy. Therefore, the total energy of the system (spring and object) remains constant.

What factors affect the amount of energy conserved in this scenario?

The amount of energy conserved in this scenario is affected by the mass of the object, the height of the incline, and the spring constant of the spring. A heavier object will have more potential and kinetic energy compared to a lighter object. A taller incline will result in a higher potential energy at the top, and a stiffer spring will store more elastic potential energy.

Is the conservation of energy always applicable in this scenario?

The conservation of energy may not always apply in this scenario if there are external factors at play, such as friction or air resistance. These external forces can cause energy to be lost from the system, resulting in a decrease in the total energy and a failure to conserve energy.

Can the potential energy of the spring be converted back into kinetic energy?

Yes, the potential energy stored in the spring can be converted back into kinetic energy. This can happen if the object on the incline is pushed back up to its original position, causing the spring to release its potential energy and propel the object up the incline. However, there will be some energy lost due to friction, resulting in a slightly lower total energy compared to the initial state.

How does the conservation of energy relate to the law of conservation of momentum in this scenario?

The conservation of energy and the law of conservation of momentum are both fundamental principles in physics. In this scenario, as the object moves down the incline and the spring compresses, the momentum of the object is conserved. However, the energy of the system is not conserved as some of the kinetic energy is transferred to the spring. This highlights the difference between the two principles and how they apply in different situations.

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