Work Done by a General Variable Force — Question 3

Question 3

The slope of accumulated work gives the local force

For a 2 kg particle moving along x, the net work accumulated from x=0 is $W(x)=6x-2x^2$ J, with x in meters. It starts at x=0 moving right at 2 m/s.

1. Find the net force as a function of x.

2. Find the position and value of the maximum speed during the first rightward motion.

3. Find the first turning position.

Question 3 - Solution

The slope of accumulated work gives the local force

(a) Differentiate with respect to position.

The governing relation is $$\boxed{F_x(x)=\frac{dW}{dx},\qquad K(x)=K_0+W(x)}$$

$$F_x=dW/dx=6-4x\ \mathrm N.$$ A work graph ordinate is energy; its slope, not its ordinate, is force.

(b) Maximize the kinetic energy.

$$K=4+6x-2x^2.$$ Its maximum occurs at x=1.5 m, where $F_x=0$ and $$K=8.5\ \mathrm J.$$ Since $K=v^2$ for this mass, $$v_{\max}=\sqrt{8.5}\ \mathrm{m/s}.$$

(c) Find the first accessible zero of K.

$$4+6x-2x^2=0$$ gives $$x=(3+\sqrt{17})/2=3.5616\ \mathrm m$$ on the first rightward trip. K remains positive before that root; the negative root lies behind the starting point.