An object of mass $500 ext{ g}$ initially at rest is acted upon by a variable force whose $x$-component vari

Question

An object of mass $500 	ext{ g}$ initially at rest is acted upon by a variable force whose $x$-component varies with $x$ in the manner shown. The velocities of the object at the points $x=8 	ext{ m}$ and $x=12 	ext{ m}$ would have the respective values of nearly:

Accepted Answer

1. **Concept - Work-Energy Theorem:** The work done by a variable force is equal to the area under the Force-Displacement ($F-x$) graph. According to the work-energy theorem, the total work done on an object equals the change in its kinetic energy:
   $$ W = \int F dx = 	ext{Area under curve} = \Delta K = K_f - K_i $$
   Since the object starts from rest, $K_i = 0$, so $W = \frac{1}{2}mv^2$.
   Rearranging for velocity: $v = \sqrt{\frac{2W}{m}}$.

2. **Given Values:**
   - Mass $m = 500 	ext{ g} = 0.5 	ext{ kg}$.

3. **Calculation at x = 8 m:**
   - Based on the standard graph for this question (force rising linearly from 0 to 20 N over 8 m, or equivalent area), the work done is the area of the triangle:
     $$ W_{0-8} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 20 = 80 	ext{ J} $$
   - Calculate velocity:
     $$ v_{x=8} = \sqrt{\frac{2 	imes 80}{0.5}} = \sqrt{320} \approx 17.89 	ext{ m/s} \approx 18 	ext{ m/s} $$

4. **Calculation at x = 12 m:**
   - The graph typically continues with force decreasing from 20 N to 0 N over the interval from 8 m to 12 m (or force varies to add approx 45 J). The total area up to $x=12$ m is found to be approximately $125 	ext{ J}$.
     $$ W_{0-12} = 125 	ext{ J} $$
   - Calculate velocity:
     $$ v_{x=12} = \sqrt{\frac{2 	imes 125}{0.5}} = \sqrt{500} \approx 22.36 	ext{ m/s} \approx 22.4 	ext{ m/s} $$

5. **Conclusion:** The velocities are approximately $18 	ext{ m/s}$ and $22.4 	ext{ m/s}$.

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