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NNADVOKAT [17]
3 years ago
13

When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp

ring?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

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Answer:

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When the displacement in the harmonic motion is maximum then kinetic energy at the maximum point is minimum and the potential energy is maximum at that point.  

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For part b)
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