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alexandr1967
14 days ago
5

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

{ J}2.5 J. If the block is replaced by a block with twice the mass of the original block and the amplitude of the motion is 6 \text{ cm}6 cm, what is the energy of the system.
Physics
1 answer:
ValentinkaMS [2.4K]14 days ago
8 0

Answer:

The system's energy amounts to 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

The spring constant needs to be computed.

Using the formula for the mechanical energy of the system,

E=\dfrac{1}{2}kA^2

Substituting the values into the formula:

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

k=\dfrac{2.5\times2}{(10\times10^{-2})^2}

k=500\ N/m

If the block is swapped out for one with double the original mass,

Amplitude = 6 cm

We need to find the energy

Using the mechanical energy formula,

E=\dfrac{1}{2}kA^2

Substituting into the formula:

E=\dfrac{1}{2}\times500\times(6\times10^{-2})

E=15\ J

Thus, the system’s energy is 15 J.

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v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

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