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olganol [36]
3 years ago
10

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

ng a further 10 cm, i.e. find the work required to stretch the spring from 10 cm beyond its natural length to 20 cm beyond its natural length. (Do not enter units)
Physics
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

10 = k\dfrac{0.1^2}{2}

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

W = \int_{0.1}^{0.2} 2000 x dx

W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

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Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

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We are told width is 20 mm and thickness 40 mm.

Thus;

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Now formula for shear modulus is;

E = σ/ε_z

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While ε_z is longitudinal strain.

Thus;

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Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

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Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

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Explanation:

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