Calculate total mechanical energy of a frictionless spring

[henrco]

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

 

[gneill]

Your method is fine.

Re-do your final calculation; you seem to have slipped a decimal point.

 

[physics user1]

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

 

[henrco]

Thank you for the feedback and for checking my shoddy calculation.

I redid the calculation and the answer came to 0.418 Joules

 

[haruspex]

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.

 

[physics user1]

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

:)