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Nostrana [21]
3 years ago
14

A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. In

itially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at 0.400 m/s in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system?
Physics
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

E_M=0.0492J.

Explanation:

The mechanical energy of the system will be the kinetic energy plus the elastic potential energy: E_M=K+U_e.

We know that the equation for the kinetic energy is K=\frac{mv^2}{2}, where <em>m </em>is the mass of the object and <em>v </em>its velocity.

We know that the equation for the elastic potential energy is U_e=\frac{k\Delta x^2}{2}, where <em>k</em> is the spring constant and \Delta x the compression (or elongation) respect to equilibrium.

So for our values we have:

E_M=K+U_e=\frac{mv^2}{2}+\frac{k \Delta x^2}{2}=\frac{(0.24kg)(0.4m/s)^2}{2}+\frac{(6N/m)(0.1m)^2}{2}=0.0492J.

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Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ v=\frac{502.1}{\sqrt{3} }

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