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Q.

The amount of work done in stretching a spring of force constant 500 N/m from a stretched length of 40 cm to 50 cm is:

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a

45 J

b

22.5 J

c

45 x 104 J

d

22.5 x 104 J

answer is B.

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Detailed Solution

Concept- Let the applied force be F and the extended length be x.
Now, F is directly proportional to x.
In the formula F = kx, k denotes the spring's force constant.
The amount of work done is defined as the product of the average force and displacement.
So, using the formula for work done that is

W = ½ k x 2

.
Let x here be x2 – x1 So,

W = ½ k x 2 – x 1 2

Given that

k = 500 N / m X 1 = 40 c m i n t o m = 0.40 m X 2 = 50 c m i n t o m = 0.50 m

So, putting these values in the formula, we get

W = ½ 500 0.50 – 0.40 2 = 22.5 J

Hence, the correct option is 2.

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The amount of work done in stretching a spring of force constant 500 N/m from a stretched length of 40 cm to 50 cm is: