What is the equation for the object's velocity after being pushed in spring?

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In summary, the conversation discusses finding the velocity and kinetic energy of an object after it is released from a spring and encounters friction. The formula for calculating the velocity is given as final velocity squared=initial velocity squared + two (acceleration)(distance), with the initial velocity being the velocity as the object leaves the spring. The kinetic energy at 4m is found to be 30J, which can be used to deduce the velocity. The conversation also addresses concerns about taking the square root of a negative number and the role of friction in the equation.
  • #1
Zero
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Homework Statement
Problem: A spring is compressed 50 cm from the equilibrium position. The spring constant is 12 N/cm. A 20 kg object is placed next to the spring initially at rest and it is released. After the object is released from the spring, the force of friction acting on the object is 30 N.

A) Calculate the force required to compress the spring by 50 cm. (Already got the answer which is 600N)

B) Calculate the energy that was stored in the spring. (Got this answer also which is 150 J)

C. Calculate the object's velocity 4m after it has left the spring. (This is where I stopped don't know what to do here)

What I tried: Unfortunately I don't know where to begin. I know what variables are which are work, spring constant, friction and potential energy in the spring which is the same as the work.

As for number 3, I'm totally lost but I do have a theory that I don't know if it would be true for this specific equation.

Theory: Would I have to find kinetic energy first.

The reason that I have such a theory is because of question D which I did not add since I did not attempt it as of yet but as of posting this, I will be attempting.
Relevant Equations
F=Kx and W=1/2FX
Et1+w=Et2
1/2mv+w=1/2mv
0+150=1/2 (20)
300=(20)v
15=v
I know this answer is wrong.
 
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  • #2
When the object leaves the spring, the energy that was stored in the spring is converted into the kinetic energy of the object. So you can calculate the speed of the object at the time when it first leaves the spring. You are told that "After the object is released from the spring, the force of friction acting on the object is 30 N." Calculate the deceleration from that.
 
  • #3
I did what would you said and got -1.5003 m/s^2 but when I use the formula final velocity squared=initial volocity+two (acceleration)(distance) I couldn't do it because I can't square root a negative number. Is the formula used wrong or am I on the right track but I just forgot a very key detail. Also, I did get the kinetic energy of the object at 4m which is 30J
 
  • #4
Zero said:
I did what would you said and got -1.5003 m/s^2 but when I use the formula final velocity squared=initial volocity+two (acceleration)(distance) I couldn't do it because I can't square root a negative number. Is the formula used wrong or am I on the right track but I just forgot a very key detail. Also, I did get the kinetic energy of the object at 4m which is 30J
Can you show your equation? Just because the velocity is decreasing and doesn't reach zero, doesn't mean that you need to take the square root of a negative number...
 
  • #5
Zero said:
final velocity squared=initial volocity+two (acceleration)(distance)
That should be
final velocity squared=initial velocity squared + two (acceleration)(distance)
What did you get for initial velocity?
Zero said:
I did get the kinetic energy of the object at 4m which is 30J
So deduce the velocity from that.
 
  • #6
haruspex said:
That should be
final velocity squared=initial velocity squared + two (acceleration)(distance)
What did you get for initial velocity?

So deduce the velocity from that.
isn't initial velocity 0 so it wouldn't matter. Yes i know the kinetic energy is 30j and the velocity is 1.73 m/s but i just don't know how to get the answer for velocity unless you did a different formula than Ek=1/2mv^2 to find kinetic energy then that is possible to deduce that way
 
  • #7
Zero said:
isn't initial velocity 0 so it wouldn't matter.
The acceleration in the equation you are using results from friction, and that does not start until the object leaves the spring. So for this equation the initial velocity is the velocity as it leaves the spring.
 
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  • #8
haruspex said:
The acceleration in the equation you are using results from friction, and that does not start until the object leaves the spring. So for this equation the initial velocity is the velocity as it leaves the spring.
Got the answer thank you all for the effort
 
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  • #9
berkeman said:
Can you show your equation? Just because the velocity is decreasing and doesn't reach zero, doesn't mean that you need to take the square root of a negative number...
Got the answer thank you all for the effort
 
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1. What is the equation for calculating an object's velocity after being pushed in a spring?

The equation for calculating an object's velocity after being pushed in a spring is v = √(2k/m) * x, where v is the velocity, k is the spring constant, m is the mass of the object, and x is the distance the spring is compressed or stretched.

2. How does the spring constant affect the object's velocity?

The spring constant, represented by k in the equation, directly affects the object's velocity. A higher spring constant will result in a higher velocity, while a lower spring constant will result in a lower velocity.

3. Can the mass of the object affect its velocity when pushed in a spring?

Yes, the mass of the object, represented by m in the equation, does affect its velocity. A heavier object will have a lower velocity compared to a lighter object when pushed in a spring with the same spring constant and distance.

4. What happens to the object's velocity if the spring is compressed or stretched further?

If the distance x in the equation is increased, the object's velocity will also increase. This means that the object will have a higher velocity if the spring is compressed or stretched further.

5. Is the equation for an object's velocity after being pushed in a spring the same for all types of springs?

No, the equation may vary depending on the type of spring being used. For example, the equation for a spiral spring may be different from the equation for a helical spring. It is important to use the correct equation for the specific type of spring being used.

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