Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# Calculus help?

A spring has a natural length of 20 cm. If a 29-N force is required to keep it stretched to a length of 26 cm, how much work W is required to stretch it from 20 cm to 23 cm? (Round your answer to two decimal places.)

W=   J

Relevance
• To simplify calculations we'll work with meters instead of centimeters. Recall there are 0.01 meters in one centimeter.

From the first bit of info given, we can calculate the spring constant k as follows:

Recall

F =  kx <------ This is the restorative force exerted by the spring when displaced a distance of x from its rest position.

In our case we have

29 = k(0.06)

Solving for k gives

k = 483.3N/m

Now, recall that work is given by

W = F(d), where F is the force applied and d is the distance displaced.

The amount of work required to displace the spring a differential distance dx for its resting position is

dW = 483.3xdx

The total work is given by

W =  ∫483.3xdx

the limits of integration are from 0 to 6

• F=kx

29=k(26-20)

k=(29/6)

W=(29/6)=14.5J