Physics, asked by krishkarpe375, 5 months ago

a spring is stretched by 50mm by applying a force the work done by the spring is. if the force required for stretching 1 mm of the spring is 10N​

Answers

Answered by tarracharan
8

Question:

A spring 50mm long spring is stretched by applying a force. If 10N force is required to stretch the spring through 1 mm, then work done in stretching the spring through 50mm is

Given:

Force (to stretch 1mm) = 10N

To Find:

Work done by stretching 50mm.

Solution:

Force constant,

\sf{k = \dfrac{F}{x} = \dfrac{10}{0.001}=10⁴ N/m}

\bold{Work\: done = \dfrac{1}{2}kx²}

\bold{ = \dfrac{1}{2}×10⁴×(\dfrac{5\cancel{0}}{100\cancel{0}})²}

\bold{ = \dfrac{1}{2}×\cancel{10⁴}×\dfrac{25}{\cancel{10⁴}}}

=\bold{\red{  12.5 J}}

Work done by stretching the spring by 50mm is 12.5J.

Answered by hotelcalifornia
2

Given:

Elongation =1mm

Force required for 1mm elongation =10N

To find:

Work done by the spring to stretch the spring by 50mm.

Solution:

Step 1

We have been given that the force required by the spring to get elongated by a length of 1mm is 10N.

According to the Hooke's law, we have, the force required to elongate a spring by displacement x is given by

F=kx

Where, k is the spring constant.

We have,  F=10N    ; x=1mm=10^{-3}m  

10=k(10^{-3} )

k=10^{4}N/m

Hence, the spring constant of the spring is 10⁴ N/m.

Step 2

Now,

We know, the work done in stretching the spring is stored as the elastic potential energy in the spring.This value is given by

E=\frac{1}{2}kx^{2}

W=\frac{1}{2}kx^{2}

We have,  k=10^{4}N/m    ; x=50mm=0.05m

Hence, the work done by the spring will be

W=\frac{1}{2}(10^{4} )(0.05)^{2}

W=\frac{25}{2}

W=12.5J

Final answer:

Hence, the work done by the spring in elongating the spring by 50 mm is 12.5 J.

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