Work calculus

Question

 A force of 7 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its natural length?

Answer 1

Hi Little,

 

Maria's answer will help you determine the force required to keep the spring stretched at 0.9ft, but not the amount of work required, as is asked in the question.

 

    Work = Force * distance

 

    W = F*x

 

This is a simple enough equation, as long as the amount of force is constant. Unfortunately, the amount of force required to compress or stretch a spring changes depending on how much it is compressed or stretched, so we will have to use integration to solve for the total amount of work.

 

Using the same equation above, each infinitely small amount of work equals Force times the infinitely small change in distance:

 

    dW = Force * dx

 

So, to find the total work required, we will integrate the expression, relative to "x" from the initial distance to the final distance:

 

    W = ∫₀^0.9 Force*dx

 

Now we need to come up with an expression for force, in terms of "x". To stretch/compress a spring, we would need to apply a different amount of force, depending on the amount it is stretched/compressed:

 

    F = k*x           where "k" is the spring constant (force/displacement) and "x" is the amount it has been stretched/compressed.

 

This makes our integral:

 

    W = ∫₀^0.9 k*x dx

 

The problem tells us that 7 lbs of force is required to maintain the spring stretched 0.3 ft, so the spring constant "k" must be (7)/(0.3) = 23.3 lbs/ft (rounded to 3 digits)

 

    W = ∫₀^0.9 23.3*x dx

 

As with any integral, we can pull the constant 23.3 out of the integral:

 

    W = 23.3 * ∫₀^0.9 x dx

 

Now let's perform the integration:

 

    W = 23.3 * (x²)/2,    evaluated from 0 to 0.9

 

    W = 23.3 * (.9²/2 - 0²/2)

 

    W = 9.44 ft*lbs (rounded to three digits)

 

I hope this helps!

 

Andy

 

Answer 2

Hi-

 

I will simply consider that if for holding a length of 0.3 feet is applied a force of 7 pounds, to bring the spring to its natural length of 0.9 feet there is to apply a greater force, and in this case it is easy to see without calculations that the number is 3; infact if we multiply 0.3 x 3, the result is 0.9, and the same 7 x 3 will be 21, and these are the numbers coming from these calculations.

 

I simply establish a proportion following this logic:

 

 

If  for L=0.3 F=7, how much will be F for L=0.9?

 

The proportion is:

 

               F1: L1 =   FX: L2

 

                 7: 0.3 = x: 0.9

 

where x stands for  the work we are looking for, and which is:

 

            x = 7 x 0.9/0.3 = 6.3/0.3 = 21 

 

The answer is 21 pounds per feet.

 

I hope this can help.