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2 years ago

14

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a sprin

g 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters

1 answer:

Bingel [31]2 years ago

5 0

Answer:

Explanation:

350 N force stretches the spring by 30 cm

spring constant K = 350 / 0.30 = (350 / 0.3) N / m

To calculate work done by a spring force we proceed as follows

spring force when the spring is stretched by x = Kx

This force is variable so work done by it can be calculated by integration

Work done by it in stretching from x₁ to x₂

W = ∫ F dx

= ∫ Kx dx with limit from x₁ to x ₂

= 1/2 K ( x₂² - x₁² )

Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m

Work done

= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )

= 227.50 J

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Answer:

+ 5 m/s

Explanation:

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3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

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Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

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Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

So

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

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From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

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This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

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$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

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3 years ago

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2 years ago

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"where crests and troughs have their maxima at the same time"

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2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

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From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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3 years ago