Vibrations & Waves: Solving Mass, Spring Constant & Energy

[Elvis]

Vibrations and waves
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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[Doc Al]

Find the total energy, just like you did for part one. Then use:

E = PE + KE
E = 1/2kA^2 + 1/2mv^2

Solve for v, when x = 5 cm.

That equation should be: E = 1/2kx^2 + 1/2mv^2. Use x, not A (which usually stands for amplitude).

 

[tomcue]

I would like to commend you for attempting to solve these problems and encourage you to continue practicing and seeking help when needed. Here is a possible solution for the second problem:

2) Using the same formula as before, we can first find the potential energy (PE) at x=5 cm:

PE = 1/2kx^2
PE = 1/2(20*100)*(0.05)^2
PE = 2.5J

Next, we can use the conservation of energy principle to find the kinetic energy (KE) at x=5 cm:

KE = E - PE
KE = 26J - 2.5J
KE = 23.5J

Finally, we can use the formula for kinetic energy to find the velocity (v):

KE = 1/2mv^2
23.5J = 1/2(15)v^2
v = √(23.5J/7.5kg)
v = 2.2 m/s (rounded to the nearest tenth)

Therefore, the speed at x=5 cm is approximately 2.2 m/s. It is important to note that the mass and spring constant are different from the first problem, so the answer will also be different. Keep up the good work and don't hesitate to ask for help when needed!