Solving Vibrations and Waves: Mass, Spring Constant, Energy, Speed

[Elvis]

1) A 5 kg mass, m, rests on a frictionless, horizontal, wooden table top, and is attached to one end of a spring anchored at its other end . If the spring has a spring constant=13 N/cm and the mass is pulled back (in the positive direction ) 20 cm and released, to the nearest tenth of a joule, what is its total energy?
2) if the mass is 15 kg and the spring constant is 20 N/cm, to the nearest tenth of a m/s, what is its speed at x= 5 cm?

I have solved the 1st one :

1N/cm=100N/m

E = PE + KE
E = 1/2kA^2 + 1/2mv^2 At KE v=0 because the mass stops momentarily as it changes directions. Then :

E = PE
E = 1/2kA^2
E = 1/2(13*100)*(0.2)^2
E = 26J

I have problem with the second problem. please help to solve this. I have 2 hours that I am tryin to solve it . I have only the answer that is 2.2m/s.

Please help......



Lol...
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[kuruman]

The spring is still pulled back 20 cm in part (b). Can you find how much energy is stored in it at the moment of release when v = 0?

 

[tomcue]

I would first like to commend you for putting in effort and attempting to solve the problem on your own. This shows your dedication and determination to understand the concepts of vibrations and waves.

To solve the second problem, we can use the same formula for total energy, E = 1/2kA^2 + 1/2mv^2, but this time we need to find the speed, v, at a specific position, x = 5 cm.

Using the given values, we can rewrite the formula as:

E = 1/2(20*100)*(0.05)^2 + 1/2(15)(v)^2

We know that at x = 5 cm, the potential energy is equal to the total energy, so we can set E = 26 J (from the first problem). This gives us:

26 J = 1/2(20*100)*(0.05)^2 + 1/2(15)(v)^2

Solving for v, we get:

v = √(2(26-10))/15 ≈ 2.2 m/s

Therefore, at x = 5 cm, the speed of the mass is approximately 2.2 m/s.

I hope this helps and clarifies any confusion you may have had. Keep up the good work!