Question

From $x=0$ to $x=6$, the force experienced by an object varies according to the function $F\left(x\right)=\sqrt{6x-x^2}$, as shown above. What is the work done by this force as the object moves from $x=0$ to $x=3$?
Assume all numbers are given in standard units.

• A. $0J$
• B. $4.5J$
• C. $7.1J$
• D. $9.0J$
• E. $14.0J$

Answer 1

Answer: (C)

$7.1J$

There is a simpler method to arrive at the solution.

Conceptually, you are integrating force over displacement, so if you are given the graph, the integration is equal to the area under the curve. The plot given is one of a semicircle.

The function $F\left(x\right)=\sqrt{6x-x^2}$ is one of a semicircle.

It may be easier to see this when it is in the form $F\left(x\right)=\sqrt{9+6x-x^2-9}$ , which corresponds to $F\left(x\right)=3^2-(x-3{)}^2$ which is the upper half of a circle centered at $x=3$ with radius $3$.

The area of a circle is given by $A=\pi r^2$, and in this case you are looking for the area under half of the top half.

Thus, $W=\frac{1}{4}A=\frac{1}{4}\pi (3{)}^2\approx 7.0658J$, that is choice "C".