Question

A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Answer 1

The work required to strecth the spring is about 2.0 Joules

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Further explanation

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

Let us now tackle the problem!

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Given:

spring constant = k = 2500 N/m

extension = x = 4.00 cm = 4.00 × 10⁻² m

Asked:

work required = W = ?

Solution:

We will calculate the work required to stretch the spring as follows:

$W = E_p$

$W = \frac{1}{2} k x^2$

$W = \frac{1}{2} \times 2500 \times ( 4.00 \times 10^{-2} )^2$

$W = 1250 \times ( 16.00 \times 10^{-4} )$

$W = 20000 \times 10^{-4}$

$W = 2.0 \times 10^4 \times 10^{-4}$

$W = 2.0 \texttt{ J}$

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Conclusion:

The work required to strecth the spring is about 2.0 Joules

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Answer details

Grade: High School

Subject: Physics

Chapter: Elasticity

Answer 2

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.