Spring/ Kinetic Energy Problem

In summary, at a waterpark, sleds with riders are propelled along a horizontal surface by a compressed spring with a force constant of 4300 N/m. The sled and rider, with a total mass of 68.0 kg, compress the spring by 0.390 m before being released with zero initial velocity. The sled's speed when the spring returns to its uncompressed length is 3.1 m/s. When the spring is still compressed by 0.225 m, the sled's speed is 2.66 m/s. This is found by considering the total energy of the system at the 0.225 m compression point.
  • #1
Elvis 123456789
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Homework Statement


At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant k = 4300 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 68.0 kg are pushed against the other end, compressing the spring 0.390 m . The sled is then released with zero initial velocity.

1.)What is the sled's speed when the spring returns to its uncompressed length?

2.)What is the sled's speed when the spring is still compressed 0.225 m ?

Homework Equations


Work done by a spring = 1/2*K*X^2

Work = change in Kinetic Energy

The Attempt at a Solution



1.) 1/2*K*X^2=K2 - K1 ------> K1=0

1/2*K*X^2=1/2*M*V2^2

V2= sqrt (K*X^2)/M = sqrt (4300*(0.39^2))/68 = 3.1 m/s

2.) I tried the same thing from the part before but instead i chose X=0.165m since that's the distance that the spring moves from the initial position when the spring is compressed 0.39m. I got 1.31 m/s which was wrong and I don't know why
 
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  • #2
Elvis 123456789 said:
instead i chose X=0.165m since that's the distance that the spring moves from the initial position
That does not work. The force decreases as the spring expands, so there is relatively more energy released in the first part than in the second.
Instead, just consider the total energy at the 0.225m of compression point.
 
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  • #3
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haruspex said:
That does not work. The force decreases as the spring expands, so there is relatively more energy released in the first part than in the second.
Instead, just consider the total energy at the 0.225m of compression point.
So then could I setup the problem as follows

1/2*K*X^2=1/2*M*V2^2 - 1/2*M*V1^2 ----> V1 = sqrt (M*V2^2 - K*X^2)/M

using V2 = 3.1 m/s
X = 0.2 m

V1= 2.66 m/s

is this correct?
 
  • #4
Elvis 123456789 said:
`

So then could I setup the problem as follows

1/2*K*X^2=1/2*M*V2^2 - 1/2*M*V1^2 ----> V1 = sqrt (M*V2^2 - K*X^2)/M

using V2 = 3.1 m/s
X = 0.2 m

V1= 2.66 m/s

is this correct?
That's the right idea, but a little inaccurate. You are given data to three significant figures. Keep 4 sig figures through all your working, and write the final answers to three.

Edit: just noticed you wrote x=.2 in your last post. The given value was .225.
 
  • #5
haruspex said:
That's the right idea, but a little inaccurate. You are given data to three significant figures. Keep 4 sig figures through all your working, and write the final answers to three.

Edit: just noticed you wrote x=.2 in your last post. The given value was .225.
oh yah, i mixed up the 0.2 for another problem I was looking at, i meant 0.255.

Do you think you can clarify a little bit more on why what i did the first time was wrong. Like why did it work for the first part and not the second.
 
  • #6
Elvis 123456789 said:
oh yah, i mixed up the 0.2 for another problem I was looking at, i meant 0.255.

Do you think you can clarify a little bit more on why what i did the first time was wrong. Like why did it work for the first part and not the second.
Consider two different extension of a spring, x and y. In your first approach you took the difference in energies as ##\frac 12k(y-x)^2##. But the energy difference is clearly ##\frac 12k(y^2-x^2)##. Are they the same?
 
  • #7
haruspex said:
Consider two different extension of a spring, x and y. In your first approach you took the difference in energies as ##\frac 12k(y-x)^2##. But the energy difference is clearly ##\frac 12k(y^2-x^2)##. Are they the same?
oh ok I see. So whenever I am to obtain the distance traveled by the spring, I take the difference of the squares with respect to the initial position?
 
  • #8
Elvis 123456789 said:
oh ok I see. So whenever I am to obtain the distance traveled by the spring, I take the difference of the squares with respect to the initial position?
For energy, yes. Not for difference in force, obviously.
 
  • #9
haruspex said:
For energy, yes. Not for difference in force, obviously.
ok I see. Thanks for the help haruspex!
 

1. What is spring/kinetic energy?

Spring/kinetic energy is a type of energy that is associated with the motion of an object. It is the energy that an object possesses due to its position or motion.

2. How is spring/kinetic energy calculated?

The formula for spring/kinetic energy is 1/2 * k * x^2, where k is the spring constant and x is the displacement of the spring. This formula can also be written as KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. What is the relationship between spring/kinetic energy and potential energy?

Spring/kinetic energy and potential energy are two forms of mechanical energy. They are interconvertible and the total mechanical energy remains constant. When an object has potential energy, it has the potential to do work and when it is in motion, it possesses kinetic energy.

4. How does the mass and velocity of an object affect its spring/kinetic energy?

The mass of an object directly affects its spring/kinetic energy. The greater the mass, the more kinetic energy it will have at a given velocity. The velocity of an object also has a direct relationship with its kinetic energy. The greater the velocity, the more kinetic energy it will have at a given mass.

5. What are some real-world examples of spring/kinetic energy?

There are many examples of spring/kinetic energy in our daily lives. A few examples include a swinging pendulum, a bouncing ball, a stretched rubber band, or a moving car. In all of these examples, energy is transferred between potential and kinetic as the object moves.

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