Question

A spring of force constant 100n/m is stretched upto 5cm. Find out work done

Answer 1

Answer: Answer is in the attachment

Explanation:

Answer 2

Answer:

The work done is 0.1250J.

Explanation:

The work done is calculated as,

$W=\frac{1}{2}kx^2$     (1)

Where,

W=work done

k=spring constant or force constant

x=streatching in the spring

From the question we have,

The force constant=100N/m

The stretching in the spring=5cm=0.05m (1cm=10⁻²m)

By inserting the entities in equation (1) we get;

$W=\frac{1}{2}\times 100\times (0.05)^2$

$W=50\times 25\times 10^-^4$

$W=0.1250J$

The work done is 0.1250J.

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