# PE and KE, mass sliding down incline into a spring (no friction)

• J-dizzal
In summary, the conversation revolves around a physics problem involving a block of mass 15 kg being released on a frictionless incline with a spring at the bottom. The problem asks for the distance the block moves down the incline before stopping at the spring, as well as the speed of the block right before it touches the spring. The conversation includes discussions on using kinematic equations and energy methods to solve for these values, as well as identifying and correcting mistakes in the calculations. Ultimately, the correct solution is found by using the energy method and equating the total energy to the kinetic energy right before the block touches the spring.
J-dizzal

## Homework Statement

In the figure, a block of mass m = 15 kg is released from rest on a frictionless incline of angle θ = 26°. Below the block is a spring that can be compressed 3.1 cm by a force of 380 N. The block momentarily stops when it compresses the spring by 4.0 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_41.gif

## The Attempt at a Solution

Last edited:
Look at your expression for Ws at the bottom... you used (0.4m)^2 instead of (0.04m)^2

Hey Nathanael, thanks. let me try this again

Nathanael said:
Look at your expression for Ws at the bottom... you used (0.4m)^2 instead of (0.04m)^2
when solving for part b I am using K=1/2 mv2 , where K=9.8065, m=15 and i get v=1.143

using kinematic eq. v2=2a(x-x0) where a=4.296 and x=0.15218. I got v=0.4461 m/s but not correct

J-dizzal said:
using kinematic eq. v2=2a(x-x0) where a=4.296 and x=0.15218. I got v=0.4461 m/s but not correct
Kinematics equations were derived with the assumption that acceleration is constant, so it doesn't apply to this situation. (Their use is pretty limited.)

J-dizzal said:
when solving for part b I am using K=1/2 mv2 , where K=9.8065, m=15 and i get v=1.143
I think this is the trap they expected you to fall into. Be careful with the energy: is there any other form of energy involved?

Nathanael said:
Kinematics equations were derived with the assumption that acceleration is constant, so it doesn't apply to this situation. (Their use is pretty limited.)I think this is the trap they expected you to fall into. Be careful with the energy: is there any other form of energy involved?
I don't see why kinematic eq won't work, because its asking for velocity before it even touches the spring. so a should be constant?
There could be a potential energy of gravity on the block. U=mgh

J-dizzal said:
I don't see why kinematic eq won't work, because its asking for velocity before it even touches the spring. so a should be constant?
You're right, I wasn't thinking about what you were doing sorry
Your mistake is that x you used is the total distance including the distance compressed. You just want the distance before it touches.

J-dizzal said:
There could be a potential energy of gravity on the block. U=mgh
Right. As the spring is compressed gravity is still doing work, so the kinetic energy of the block right before it touches is not equal to the total energy of the system.

Nathanael said:
You're right, I wasn't thinking about what you were doing sorry
Your mistake is that x you used is the total distance including the distance compressed. You just want the distance before it touches.Right. As the spring is compressed gravity is still doing work, so the kinetic energy of the block right before it touches is not equal to the total energy of the system.
wouldnt the total distance be .15218+.04?

Edit. nevermind its .15218-.04 for the dist before spring

Edit2. after plugging in .11218 i still am doing something wrong when solving for v

Edit3. ok i got it. i was trying to square .11218 in the kinematic equation, i don't know why i did that.

Last edited:
Since you got it I will tell you the energy method of doing it: ##E_{kinetic}=E_{total}-0.04mg\sin(26°)## and you already determined Etotal for an earlier part.

What you used in post #4 was Ekinetic=Etotal

J-dizzal

## 1. What is the relationship between potential energy and kinetic energy?

The potential energy (PE) and kinetic energy (KE) of a system are interrelated. As an object loses potential energy, it gains kinetic energy, and vice versa. This is known as the law of conservation of energy.

## 2. How does mass affect the PE and KE of an object sliding down an incline?

The mass of an object does not affect its potential or kinetic energy in this scenario. As long as the incline and the spring are unchanged, the object's mass will not impact its energy.

## 3. Does friction play a role in the PE and KE of a sliding object?

In this particular scenario, friction is not a factor, so it does not affect the object's potential or kinetic energy. However, in real-world situations, friction can significantly impact the energy of an object.

## 4. How does the height of the incline affect the PE and KE of the sliding object?

The higher the incline, the greater the potential energy of the object at the top. As the object slides down the incline, its potential energy decreases, and its kinetic energy increases. The height of the incline affects the amount of potential energy the object has at the top, but it does not affect the total energy of the system.

## 5. What happens to the energy of the system when the object reaches the spring at the bottom of the incline?

When the object reaches the spring, it has lost all of its potential energy and has gained an equal amount of kinetic energy. The spring then stores this kinetic energy as potential energy, causing the object to bounce back up the incline. This conversion of energy between potential and kinetic continues until the object comes to a rest at the bottom of the incline.

• Introductory Physics Homework Help
Replies
27
Views
6K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
2K