# Use Hooke's Law to determine the variable force in the spring problem. A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 cm to 50 cm?

As per Hooke's Law, the force needed to stretch/compress a spring x units from its natural length is

`F = kx`

where F is the force in Newtons (N), x is the displacement of the object from its equilibrium position (x=0) in meters and k is the spring constant.

To...

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As per Hooke's Law, the force needed to stretch/compress a spring x units from its natural length is

`F = kx`

where F is the force in Newtons (N), x is the displacement of the object from its equilibrium position (x=0) in meters and k is the spring constant.

To determine the variable force using the Hooke's Law, the spring constant should be solved. To do so, plug-in F = 250N and x = 0.30m.

`250 = k * 0.30`

`250/0.30=k`

`2500/3=k`

Plugging this to the formula of Hooke's Law, the force needed to stretch the spring x units from its natural length is

`F = 2500/3x`

Now that the expression that represents the variable force is known, let's determine the amount of work done in stretching the spring from 20 cm to 50cm.

Take note that if the force applied is constant in moving an object from x=a to x=b, the formula is

`W =F * Delta x`

However, if the force is not constant, the amount of work done in moving an object from x=a to x=b is

`W = int_a^b Fdx`

Applying this formula, the integral needed to compute the work done in stretching from x=20 cm to 50 cm is:

`W = int_0.2^0.5 2500/3x dx`       (Take note that the x should be in meters.)

`W = 2500/3 int_0.2^0.5 xdx`

To take the integral of this, apply the formula `int x^n dx = x^(n+1)/(n+1)` .

`W = 2500/3 *x^2/2`  `|_0.2^0.5`

`W = (1250x^2)/3` `|_0.2^0.5`

`W = (1250*0.5^2)/3 - (1250*0.2^2)/3`

`W=87.5`

Therefore, the amount of work done in stretching the spring from 20cm to 50 cm is 87.5 Joules.

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