1. **State the problem:** We need to find the net accelerating force $F$ required to accelerate a machine table of mass $m=540$ kg from rest to a velocity $v=0.53$ m/s over a distance $d=200$ mm (which is 0.2 m). 2. **Relevant formula:** The net force can be found using Newton's second law: $$F = m a$$ where $a$ is the acceleration. 3. **Find acceleration $a$:** Since the table starts from rest and reaches velocity $v$ over distance $d$, use the kinematic equation: $$v^2 = 2 a d$$ Solving for $a$: $$a = \frac{v^2}{2 d}$$ 4. **Calculate acceleration:** $$a = \frac{(0.53)^2}{2 \times 0.2} = \frac{0.2809}{0.4} = 0.70225 \text{ m/s}^2$$ 5. **Calculate net force $F$:** $$F = m a = 540 \times 0.70225 = 379.215 \text{ N}$$ 6. **Round to appropriate significant figures:** $$F \approx 378 \text{ N}$$ **Answer:** The net accelerating force required is approximately **378 N**.