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3 years ago

6

A 0.l ‑kilogram block is attached to an initially unstretched spring of force constant k = 40 N/m as shown right. The block is d

isplaced 0.1 m from the equilibrium point and released from rest at time t = 0.
What is the maximum potential energy of the oscillating system?

40 J

0.1 J

0.1 J

0.2 J

1 answer:

GalinKa [24]3 years ago

8 0

Answer:

The maximum potential energy of the system is 0.2 J

Explanation:

Hi there!

When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.

The equation of elastic potential energy (EPE) is the following:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretching distance.

The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.

Then:

EPE = 1/2 · 40 N/m · (0.1 m)²

EPE = 0.2 J

The maximum potential energy of the system is 0.2 J

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