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
[tex]\int dW = \int_0^{0.1} k.x dx[/tex]
[tex]W = \int_0^{0.1} k.x dx[/tex]
[tex]W = k[\dfrac{x^2}{2}]_0^{0.1}[/tex]
[tex]10 = k\dfrac{0.1^2}{2}[/tex]
k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.
[tex]W = \int_{0.1}^{0.2} 2000 x dx[/tex]
[tex]W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx[/tex]
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
The work required to produce the new extension of the spring is 40 J.
The given parameters;
The elastic constant of the spring is calculated as follows;
[tex]E = \frac{1}{2} kx^2\\\\2E = kx^2\\\\k = \frac{2E}{x^2} \\\\k = \frac{2\times 10}{(0.1)^2} \\\\k = 2,000 \ N/m[/tex]
The new extension of the spring is 20 cm. The work required to produce this extension is calculated as follows;
[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 2,000 \times (0.2)^2\\\\E = 40 \ J[/tex]
Thus, the work required to produce the new extension of the spring is 40 J.
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