Work — Question 3

Question 3 – Solution

1. Identify the work model.

The force is positive over $0\le x\le2$, so the work equals the area under the force curve:$$W=\int_0^2F(x)\,dx.$$

2. Substitute the force.

$$W=\int_0^2\frac{10}{1+x^2}\,dx=10\int_0^2\frac1{1+x^2}\,dx.$$

3. Use the inverse-tangent antiderivative.

Since $\dfrac{d}{dx}\arctan x=\dfrac1{1+x^2}$,$$W=10\left[\arctan x\right]_0^2.$$Inverse-trigonometric values are in radians.

4. Apply the endpoints.

$$W=10(\arctan2-\arctan0)=10\arctan2.$$

5. State the exact and approximate work.

$$\boxed{W=10\arctan2\ \mathrm{J}}\approx11.0715\ \mathrm{J}.$$The force decreases from $10$ N to $2$ N; the resulting work lies between $2(2)=4$ J and $10(2)=20$ J.