A force of 18 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length

W = 36 lb-in

Explanation:

Given:

Force required to pull a spring (F) = 18 lb

Elongation in the spring (x₁) = 4 in

New elongation (x₂) = 6 in

We know that, the force required to stretch or compress the spring is given by the formula:

Express in terms of 'k'. This gives,

Now, plug in 'F' and 'x₁' values and solve for 'k'. This gives,

Now, work done in stretching or compressing a spring by a length of 'x' is given as:

Here, . Solve for work, 'W'. This gives,

Therefore, the work done in stretching it from its natural length to 6 inch beyond its natural length is 36 lb-in

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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