Work and Kinetic Energy — Question 10

Question 10

A stopping-distance bound must include the work before braking

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 1000 kg vehicle travels at 20 m/s. It continues at that speed for a 0.50 s reaction interval, then brakes with a constant net opposing force of magnitude 5000 N. Ignore grade and other resistance.

1. Find braking distance and total stopping distance.

2. Is an obstacle 48 m ahead avoidable under this model?

3. Find the minimum constant net braking force needed to stop within 48 m with the same reaction interval.

Question 10 - Solution

A stopping-distance bound must include the work before braking

(a) Separate the two physical phases.

The governing relation is $$\boxed{d_{\mathrm{total}}=v_0t_r+\frac{mv_0^2}{2F_b}}$$ Initial kinetic energy is 200000 J. Braking distance is $$200000/5000=40\ \mathrm m;$$ reaction travel is $$20(0.50)=10\ \mathrm m.$$ Total stopping distance is 50 m.

(b) Compare with available distance.

No. Only 38 m remains for braking after the reaction phase, less than the required 40 m. The supplied model predicts contact before rest.

(c) Use the remaining work distance.

Required opposing force satisfies $$F(38)\ge200000,$$ giving $$F\ge100000/19\approx5263.16\ \mathrm N.$$ Using all 48 m for braking would wrongly omit the reaction interval.