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1 year ago

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A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e

nergy of the spring?

1 answer:

SVEN [57.7K]1 year ago

8 0

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

We have to find the spring constant and potential energy.

The spring constant can be calculated by the formula

$\begin{gathered} F=\text{ kx} \\ mg\text{ = kx} \\ k\text{ = }\frac{mg}{x} \end{gathered}$

Here, k is the spring constant.

g = 9.8 m/s^2 is the acceleration due to gravity.

Substituting the values, the spring constant will be

$\begin{gathered} k=\frac{0.2\times9.8}{0.03} \\ =\text{ 65.33 N/m } \end{gathered}$

The potential energy can be calculated as

$\begin{gathered} U=\frac{1}{2}kx^2 \\ =\frac{1}{2}\times65.33\text{ }\times(0.03)^2 \\ =\text{ 0.0293 J} \end{gathered}$

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3 years ago

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a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

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W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

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To learn more about work done click on the given link brainly.com/question/25573309

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1 year ago

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Answer:

Explanation:

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$\frac{r}{z}=tan(45)$

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and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

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So the Langrangian is given by:

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Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

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