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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A particle moves under the effect of a force F=CxF = Cx from x=0x = 0 to x=x1x = x_1. The work done in the process is:

A

Cx12C x_1^2

B

12Cx12\frac{1}{2} C x_1^2

C

Cx1C x_1

D

Zero

Step-by-Step Solution

The work done by a variable one-dimensional force F(x)F(x) in displacing a particle from an initial position xix_i to a final position xfx_f is given by the definite integral of the force with respect to displacement: W=xixfF(x)dxW = \int_{x_i}^{x_f} F(x) \, dx

Given that the force is F=CxF = Cx, the initial position is xi=0x_i = 0, and the final position is xf=x1x_f = x_1. Substituting these values into the formula, we get: W=0x1CxdxW = \int_{0}^{x_1} Cx \, dx W=C[x22]0x1W = C \left[ \frac{x^2}{2} \right]_{0}^{x_1} W=C(x122022)=12Cx12W = C \left( \frac{x_1^2}{2} - \frac{0^2}{2} \right) = \frac{1}{2} C x_1^2

Thus, the work done in the process is 12Cx12\frac{1}{2} C x_1^2.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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