Elastic potential energy (spring) problem

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A 0.200 kg object attached to a spring with a spring constant of 77.0 N/m is compressed by a 10.0 N force, resulting in a compression of 0.13 m and work done of 0.649 J. The maximum speed of the object after the force is released is calculated to be 2.54 m/s, occurring when the spring's potential energy is zero. For the spring's position when the object's speed is 1 m/s, the correct approach involves using energy conservation principles, leading to a position of 0.051 m. The discussion emphasizes the importance of specifying conditions for maximum speed and correctly applying energy conservation in calculations. Overall, the thread provides clarity on solving spring-related energy problems.
newguy_13
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Homework Statement


A 0.200 kg object is attached to a horizontal spring with a spring constant of 77.0 N/m. The other end of the spring is attached to a wall in such a way that it rests on a frictionless horizontal surface. A 10.0 N force is exerted on the spring, causing it to compress

a) How far will the spring be compressed?
b) how much work is done on the spring to compress it?
c) Once the force on the spring is released, the spring and object are set into harmonic motion. What will the maximum speed of the object be? Explain where this speed will occur. (no friction and mass of spring is negligible)
d) What will spring's position be when v=1m/s?

Homework Equations


Ee=1/2kx^2

The Attempt at a Solution


a)
f=kx
x=f/k
x=0.13m
b)
Ee=1/2kx^2
=1/2(77)(0.13)^2
=0.649J
c)
1/2mv^2=1/2kx^2
1/2(0.2)v^2 = 0.649
v=2.54m/s
d)
1/2mv^2 = 1/2kx^2
0.1=1/2(77)x^2
2(0.1)/77=x^2
0.051m=x

Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2

Can someone please help me? Thank you very much
 
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newguy_13 said:
Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2
That's the correct approach: energy is conserved. Your mistake in part d was setting the KE (when v=1m/s) equal to the total energy.

In part c you got lucky since when the KE is max the PE is zero. (Where does that happen?)
 
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newguy_13 said:

Homework Statement


A 0.200 kg object is attached to a horizontal spring with a spring constant of 77.0 N/m. The other end of the spring is attached to a wall in such a way that it rests on a frictionless horizontal surface. A 10.0 N force is exerted on the spring, causing it to compress

a) How far will the spring be compressed?
b) how much work is done on the spring to compress it?
c) Once the force on the spring is released, the spring and object are set into harmonic motion. What will the maximum speed of the object be? Explain where this speed will occur. (no friction and mass of spring is negligible)
d) What will spring's position be when v=1m/s?

Homework Equations


Ee=1/2kx^2

The Attempt at a Solution


a)
f=kx
x=f/k
x=0.13m
b)
Ee=1/2kx^2
=1/2(77)(0.13)^2
=0.649J
c)
1/2mv^2=1/2kx^2
1/2(0.2)v^2 = 0.649
v=2.54m/s
d)
1/2mv^2 = 1/2kx^2
0.1=1/2(77)x^2
2(0.1)/77=x^2
0.051m=x

Im not sure if for the last 2 questions I should have it as:

0.649=Epfinal+Ekfinal
0.649=1/2kx^2+1/2mv^2

Can someone please help me? Thank you very much
Hello @newguy_13,

Welcome to PF! :welcome:

a) through c) look good to me with the possible exceptions of minor rounding errors.

Don't forget that c) requires you to specify where the maximum speed occurs. (You forgot to specify the location.)

For d), use your second approach. Your first approach is not correct.

It turns out that you already used your second approach when doing part c), even if you didn't realize it. (At the point of maximum speed, what is the system's potential energy? :wink:)
 
Thanks so much for the help guys!
 
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