Section 5.4 Work
work was done upon the object.Subsection 5.4.1 Work with Constant Force
If a constant of magnitude F is applied in the direction of motion of an object and if that object moves a distance d then we define the work W performed on the object by the force to be \(W=F \cdot d\text{.}\)
Example 5.4.1.
How much work is done in lifting a 5 lbs. book from a height of 3 ft. to 7 ft.?
| system | force | distance | work |
| SI | newtons (N) | meters (m) | joules (J) |
| BE | pounds (lbs) | feet (ft) | footpounds (ftlb) |
| * 1 pound = 4.45 newtons | |||
Subsection 5.4.2 Work with Variable Force
Suppose an object moves in the positive direction along a coordinate line while subjected to a variable force f(x) that is applied in the direction of motion over the interval [a,b]. Then we define the work W performed by the force F on the object to be \(W = \displaystyle \int_a^b f(x)dx \)
Hooke's Law.
A spring that is stretched x units beyond its natural length pulls back with a force of \(F = k \cdot x\text{,}\) where k is a constant called the "spring constant". The value of k depends on such factors as the thickness of the spring, material, etc.
Worksheet Spring Problem
A spring exerts a force of 30 N when stretched 2 m beyond its natural length.
Find the spring constant k then find the force exerted by the spring when stretched 5 meters beyond its natural length. Solution
\begin{align*} f(x) \amp = k \cdot x\\ f(2) \amp = 30\\ f(2) \amp = k \cdot 2\\ 2k \amp = 30\\ k \amp = 15 \frac{N}{m} \end{align*}\begin{align*} f(x) \amp = 15x\\ f(5) \amp = 15(5)\\ \amp = 75 N \end{align*}Find the work that is done by stretching the spring form rest to 5 meters beyond its natural length. Solution
\begin{align*} W \amp = \int_0^5 15x dx\\ \amp = \left[ \dfrac{15x^2}{2} \right]_0^5\\ \amp = \dfrac{15(5)^2}{2} -\dfrac{15(0)^2}{2}\\ \amp = \dfrac{375}{2}\\ \amp = 187.5 J \end{align*}