Work and Kinetic Energy — Question 2

Question 2

A turning point limits a forward displacement calculation

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 1 kg particle starts at x=0 with velocity +4 m/s. A constant net force $-2$ N acts along x for all subsequent times.

1. Find $K(x)$ on the initial forward motion and the turning position.

2. Can the particle first reach x=5 m under these conditions?

3. After reversal, what is its velocity on returning to x=0?

Question 2 - Solution

A turning point limits a forward displacement calculation

(a) Integrate the constant force over displacement.

The governing relation is $$\boxed{K(x)=K_i+\int_0^xF_{\mathrm{net}}\,dx\geq0}$$

$$K_i=8\ \mathrm J,\qquad W= -2x\ \mathrm J$$ with x in metres. Thus $$K(x)=8-2x$$ and the forward motion ends at $$x=4\ \mathrm m$$ where $K=0$.

(b) Require nonnegative kinetic energy along the path.

At x=5 m the formula would give $$K=-2\ \mathrm J,$$ which is impossible. The particle turns at 4 m; negative kinetic energy is not an instruction to take a negative speed.

(c) Use energy and the established direction.

At x=0, K returns to 8 J, so speed is 4 m/s. The continuing negative force has made the particle move left: velocity is $-4\ \mathrm{m/s}$. Energy fixes the magnitude; the motion history fixes the sign.