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4 years ago

15

A mass on a spring will oscillate vertically when it is lifted and released how will the values of elastic potential energy and

gravitational potential energy vary during the oscillation

1 answer:

Rama09 [41]4 years ago

6 0

The answer to this is

<span>A mass on a spring will oscillate vertically when it is lifted and released. The gravitational potential energy increases from a minimum at the lowest point to a maximum at the highest point. ... The total mechanical energy at those points is the sum of the elastic and gravitational potential energies.</span>

Hope This Helps!
:D

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$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

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$\theta=tan^{-1}(-1.2)$

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$\theta=360-50.2$

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Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

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$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

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