Work Done by a General Variable Force — Question 10

Question 10

A varying drag force removes energy over a finite travel distance

A 1 kg slider enters a rough region at x=0 with speed 6 m/s. While moving right, the only horizontal force is $F_x=-(2+2x)$ N, with x in meters. The region extends to x=4 m.

1. Find kinetic energy as a function of x during rightward travel.

2. Determine whether the slider reaches the far edge.

3. Find the work done up to its first stop and explain why extending the expression beyond that point is not physical.

Question 10 - Solution

A varying drag force removes energy over a finite travel distance

(a) Integrate the negative force.

The governing relation is $$\boxed{K(x)=\tfrac12mv_0^2+\int_0^xF_x(s)\,ds\geq0}$$

$$W(x)=-2x-x^2\ \mathrm J,$$ so $$K(x)=18-2x-x^2\ \mathrm J$$ while rightward motion persists.

(b) Locate the first zero before the edge.

$$x_s=-1+\sqrt{19}=3.3589\ \mathrm m,$$ less than 4 m. Thus it stops inside the rough region; the formal value $$K(4)=-6\ \mathrm J$$ diagnoses an inaccessible endpoint.

(c) Stop the motion-dependent model at its boundary.

The work to the stop is $-18\ \mathrm J$. The law was specified for rightward sliding only. Subsequent static behavior or a reversed-motion force requires additional information, and negative kinetic energy has no physical interpretation.