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3 years ago

Times will the kinetic one

Physics

1 answer:

mixer [17]3 years ago

3 0

Answer:

The new kinetic energy is 9 times of the initial kinetic energy.

Explanation:

The kinetic energy of an object is given by the formula as follows :

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

Where

m is mass and v is speed of the body

If the speed of the body is tripled. Let v' is the new speed, v' = 3v. The new kinetic energy is E' and it is given by :

$E'=\dfrac{1}{2}mv'^2\\\\\text{Put v' = 3v}\\\\E'=\dfrac{1}{2}m\times (3v)^2\\\\=\dfrac{1}{2}m\times 9v^2\\\\=9\times (\dfrac{1}{2}mv^2)\\\\E'=9\times E$

Hence, the new kinetic energy is 9 times of the initial kinetic energy.

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3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

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

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

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3 years ago