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.)
- DerealizationLv 52 months ago
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:
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
- no sea naboLv 62 months ago