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A 0.250 kg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/m. When the toy is 0.0160 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.
A) What is the toy's total energy at any point of its motion?
B) What is the toy's amplitude of the motion?
C) What is the toy's maximum speed during its motion?

Sagot :

xero099

Answer:

a) The total energy of the toy is 0.0584 joules.

b) The maximum amplitude of the toy is approximately 0.0197 meters.

c) The maximum speed during the motion of the toy is approximately 0.684 meters per second.

Explanation:

This exercise can be solved easily by means of the formulae for Simple Harmonic Motion and Principle of Energy Conservation.

a) The total energy of the toy is the sum of elastic potential energy and translational kinetic energy, both in joules:

[tex]E = \frac{1}{2}\cdot k \cdot A^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]E[/tex] - Total energy, in joules.

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]A[/tex] - Amplitude, in meters.

[tex]m[/tex] - Mass, in kilograms.

[tex]v[/tex] - Speed, in meters per second.

If we know that [tex]k = 300\,\frac{N}{m}[/tex], [tex]A = 0.016\,m[/tex], [tex]m = 0.250\,kg[/tex] and [tex]v = 0.4\,\frac{m}{s}[/tex], then the total energy of the system is:

[tex]E = \frac{1}{2}\cdot k \cdot A^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]E = \frac{1}{2}\cdot \left(300\,\frac{N}{m} \right)\cdot (0.016\,m)^{2} + \frac{1}{2}\cdot (0.250\,kg)\cdot \left(0.4\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 0.0584\,J[/tex]

The total energy of the toy is 0.0584 joules.

b) The maximum amplitude occurs when translational kinetic energy is zero. Hence, we can determine it by this expression:

[tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex] (2)

If we know that [tex]E = 0.0584\,J[/tex] and [tex]k = 300\,\frac{N}{m}[/tex], then the maximum amplitude of the system is:

[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]

[tex]A = \sqrt{\frac{2\cdot (0.0584\,J)}{300\,\frac{N}{m} } }[/tex]

[tex]A \approx 0.0197\,m[/tex]

The maximum amplitude of the toy is approximately 0.0197 meters.

c) The maximum speed occurs when elastic potential energy is zero. Hence, we can determine it by this expression:

[tex]E = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (3)

If we know that [tex]E = 0.0584\,J[/tex] and [tex]m = 0.250\,kg[/tex], then the maximum speed during the motion is:

[tex]v = \sqrt{\frac{2\cdot E}{m} }[/tex]

[tex]v = \sqrt{\frac{2\cdot (0.0584\,J)}{0.250\,kg} }[/tex]

[tex]v \approx 0.684\,\frac{m}{s}[/tex]

The maximum speed during the motion of the toy is approximately 0.684 meters per second.

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