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2 years ago

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A chain of length L has a mass of m which is uniformly distributed. When it is placed on a smooth horizontal table, 1/4 of its l

ength hangs over the edge of the table (as shown in the figure). Then the chain slides away from the table freely, how much gravitational potential energy has changed from the beginning to the chain just sliding away from the table? (The height of the table from the ground is much greater than L)

1 answer:

krok68 [10]2 years ago

6 0

Gravitational potential energy has changed from the beginning to the chain just sliding away from the table willl be MgL/24.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

Length of hanging part, L/4

The gravitational potential energy is;

Work = Gravitational potential energy

Work done =force×displacement

$\rm W = \frac{Mg}{3} \times \frac{L}{8} \\\\ W = \frac{MgL}{24}$

Hence, gravitational potential energy has changed from the beginning to the chain just sliding away from the table willl be MgL/24.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855

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$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

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Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

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$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

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$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

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