Question

A force of 7.5 stretches a certain spring by 5cm. how much work is done in stretching this spring by 8.0 cm

Answer 1

Given :

▪ A force of 7.5N stretches a certain spring by 5cm.

To Find :

▪ Work done in stretching this spring by 8cm.

Concept :

↗ First we have to find out spring constant of the spring.

↗ If spring is elongated by an amount x then work done by the spring force on the block is given by

$\bigstar\bf\:W=\dfrac{1}{2}kx^2$

Calculation :

$\implies\sf\:F=kx_1\\ \\ \implies\sf\:7.5=k(0.05)\\ \\ \implies\sf\:k=\dfrac{7.5}{0.05}\\ \\ \implies\bf\:k=150\:Nm^{-1}\\ \\ \dashrightarrow\sf\:W=\dfrac{1}{2}k{x_2}^2\\ \\ \dashrightarrow\sf\:W=\dfrac{1}{2}(150)(0.08)^2\\ \\ \dashrightarrow\underline{\underline{\bf{W=0.48J}}}$

Answer 2

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$\huge\tt{GIVEN:}$

• A force of 7.5N streches a spring by 5 cm

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$\huge\tt{TO~FIND:}$

• So ,what force would be needed to stretch it to 8 cm?

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$\huge\tt{CONCEPT~USED:}$

As in these kind of questions, we have to find out the constant stretch of the spring,

If we say that the spring is elongated by an amount then the force or the work might be like,

• W = ½kx²

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$\huge\tt{SOLVING:}$

So, putting the values here,

⇝F = kx_1

⇝7.5 = k × 0.05

⇝7.5/0.05 = k

⇝150 Nm-¹ = K

Then,

⇝W = ½kx²

⇝W = ½ × 150 × 0.08²

$\Huge\tt\gray{⇝W = 0.48 J}$

$\rule{200}3$