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3 years ago

7

A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

e ground?

1 answer:

Inessa05 [86]3 years ago

3 0

initially coin is at rest and then it drop for total time t = 1.5 s

so here the speed of the coin at which it will hit the floor is to be find

$v_f = v_i + at$

here we know that

$v_i = 0$

a = 9.8 m/s^2

t = 1.5 s

now from above equation

$v_f = 0 + 1.5 (9.8)$

$v_f = 14.7 m/s$

so it will hit the floor with speed 14.7 m/s

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Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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Given that,

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Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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