Calculate total mechanical energy of a frictionless spring

  • Thread starter henrco
  • Start date
  • #1
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Hi,
Could you please help me determine if I've worked this problem out correctly.

1. Homework Statement


A 2.86-kg object on a frictionless horizontal surface oscillates at the end of a spring with an amplitude of
7.81 cm. Its maximum acceleration is 3.74 m/s2.
Calculate the total mechanical energy.

Homework Equations



1) Total Mechanical Energy: E = 0.5 k A^2
2) Spring Constant: k = m w^2 (where w = angular velocity)
3) Acceleration : a = -w^2 x

The Attempt at a Solution



Given the Amplitude A, we must calculate the spring constant k to calc the total mechanical energy.

First find w using formula 3)
w = - sqr( a / x)
w = - sqr(3.74 / 0.0781)
w = - 6.92

Now substitute k in formula 1 with formula 2 to calculate the total mechanical energy.

E = 0.5 (m w^2) A^2
E= 0.5 (2.86) (-6.92)^2 (0.0781)^2
E= 0.0418 Joules
 

Answers and Replies

  • #2
gneill
Mentor
20,874
2,837
Your method is fine.

Re-do your final calculation; you seem to have slipped a decimal point.
 
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Likes henrco
  • #3
210
15
Hi,
Could you please help me determine if I've worked this problem out correctly.

1. Homework Statement


A 2.86-kg object on a frictionless horizontal surface oscillates at the end of a spring with an amplitude of
7.81 cm. Its maximum acceleration is 3.74 m/s2.
Calculate the total mechanical energy.

Homework Equations



1) Total Mechanical Energy: E = 0.5 k A^2
2) Spring Constant: k = m w^2 (where w = angular velocity)
3) Acceleration : a = -w^2 x

The Attempt at a Solution



Given the Amplitude A, we must calculate the spring constant k to calc the total mechanical energy.

First find w using formula 3)
w = - sqr( a / x)
w = - sqr(3.74 / 0.0781)
w = - 6.92

Now substitute k in formula 1 with formula 2 to calculate the total mechanical energy.

E = 0.5 (m w^2) A^2
E= 0.5 (2.86) (-6.92)^2 (0.0781)^2
E= 0.0418 Joules
Yes it's okay but pay attention at this:

a= -ω^2 x ---> ω= √(-a/x)

Since a is opposite to x a= -3.74 m/s^2 and ω= 6.92 rad/s so ω>0
 
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  • #4
47
2
Thank you for the feedback and for checking my shoddy calculation.

I redid the calculation and the answer came to 0.418 Joules
 
  • #5
haruspex
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Yes it's okay but pay attention at this:

a= -ω^2 x ---> ω= √(-a/x)

Since a is opposite to x a= -3.74 m/s^2 and ω= 6.92 rad/s so ω>0
You mean, so ω is real. Whether you choose to take the positive or negative root is another matter.
 
  • #6
210
15
You mean, so ω is real. Whether you choose to take the positive or negative root is another matter.
Yes, I just focused on the fact that he got a ω< 0 by making a little mistake in solving a= -ω^2 x

:)
 

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