Speed when Spring is Stretched

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Homework Help Overview

The discussion revolves around a physics problem involving a block attached to a spring, where a constant force is applied to stretch the spring. The participants are exploring concepts related to energy conservation, work done by forces, and the relationship between potential energy in the spring and kinetic energy of the block.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning how the applied force influences the system's energy. There are attempts to rearrange energy equations to solve for speed, while others emphasize the importance of considering the work done by the applied force.

Discussion Status

There is an ongoing exploration of the correct approach to relate the energy in the system to the applied force and the resulting motion. Some participants have offered hints about the energy equation and the need to clarify the role of the applied force, while others are still grappling with the correct interpretation of the problem.

Contextual Notes

Participants are working under the assumption that the table is frictionless and are discussing the implications of the applied force on the energy of the system. There is a noted confusion regarding the correct displacement to use in the energy equations.

Manni
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A 1.50 kg block is at rest on a table and is attached to a horizontal spring with a spring constant of 19.6 N/m. The spring is initially not stretched. A constant force of 20 N horizontal force is applied to the object causing the spring to stretch. Determine the speed of the block when it has stretched 0.300 m. Assume the table is frictionless.

The way I did it:

E = (mv^2)/2 + (kx^2)/2
1/2kA^2 = (mv^2)/2 + (kx^2)/2

Do I simply rearrange for speed?
 
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What about the 20 N force? Where does that come in?

(But considering energy is the right idea.)
 
Doc Al said:
What about the 20 N force? Where does that come in?

(But considering energy is the right idea.)

I use it to find the amplitude

F = -kx
20 = -kA
-A = 20/k

Negative sign can be omitted since amplitude is treated in a A/-A fashion.
 
Manni said:
I use it to find the amplitude

F = -kx
20 = -kA
-A = 20/k
No, that's not quite right. The force is continuously applied. Instead, consider the work done by that force.
 
Yes, you can rearrange for v. :D
 
Great, but when I did that I still got the wrong answer.

The answers I have to choose from are:

4.47 m/s, 70.6 m/s, 44.7 m/s, 2.61 m/s, and 8.96 m/s
 
Last edited:
Another hint. You'll use this:
Manni said:
E = (mv^2)/2 + (kx^2)/2
But what's E? What's putting energy into this system?
 
what's your wrong answer? and what's the right one?
 
maCrobo said:
what's your wrong answer? and what's the right one?

I got 3.52 m/s, and it's not even offered as one of the choices.
 
  • #10
maCrobo said:
Yes, you can rearrange for v. :D
No you can't.
 
  • #11
Then, I'm getting curious. E = (mv^2)/2 + (kx^2)/2 can't he simply rearrange for v? O_O why?
 
  • #12
maCrobo said:
Then, I'm getting curious. E = (mv^2)/2 + (kx^2)/2 can't he simply rearrange for v? O_O why?
Before solving for v, you must first find the correct expression for the energy, E.
 
  • #13
Think about this: the spring and the mass can be thought as a system, then you apply a force F on it... so what happens to the energy?
 
  • #14
Work energy theorem seems like the simplest way to solve it.
 
  • #15
Oh, I got it!

You need to sub in (A-0.300) meters for x, and not 0.300 m!
 
  • #16
Manni said:
Oh, I got it!

You need to sub in (A-0.300) meters for x, and not 0.300 m!
:confused:

At the point in question, the displacement from equilibrium x is given as 0.300 m. (Forget about the amplitude; this is not SHM.)
 
  • #17
Elastic P.E + K.E = Total Energy = Work done when the mass travels 0.3m under a force whose magnitude is 20N
 

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