Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

An object of mass $500 \text{ g}$ 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 $x=8 \text{ m}$ and $x=12 \text{ m}$ would have the respective values of nearly:

18 m/s and 22.4 m/s

23 m/s and 22.4 m/s

23 m/s and 20.6 m/s

18 m/s and 20.6 m/s

Step-by-Step Solution

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: $W = \int F dx = \text{Area under curve} = \Delta K = K_f - K_i$ Since the object starts from rest, $K_i = 0$ , so $W = \frac{1}{2}mv^2$ . Rearranging for velocity: $v = \sqrt{\frac{2W}{m}}$ .
Given Values:

Mass $m = 500 \text{ g} = 0.5 \text{ kg}$ .

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: $W_{0-8} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 20 = 80 \text{ J}$
Calculate velocity: $v_{x=8} = \sqrt{\frac{2 \times 80}{0.5}} = \sqrt{320} \approx 17.89 \text{ m/s} \approx 18 \text{ m/s}$

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 $125 \text{ J}$ . $W_{0-12} = 125 \text{ J}$
Calculate velocity: $v_{x=12} = \sqrt{\frac{2 \times 125}{0.5}} = \sqrt{500} \approx 22.36 \text{ m/s} \approx 22.4 \text{ m/s}$

Conclusion: The velocities are approximately $18 \text{ m/s}$ and $22.4 \text{ m/s}$ .

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