Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block of mass $10 \text{ kg}$ , moving in the $x$ -direction with a constant speed of $10 \text{ m/s}$ , is subjected to a retarding force $F = 0.1x \text{ J/m}$ during its travel from $x = 20 \text{ m}$ to $30 \text{ m}$ . Its final kinetic energy will be:

475 J

450 J

275 J

250 J

Step-by-Step Solution

Calculate Initial Kinetic Energy ( $K_i$ ): Given mass $m = 10 \text{ kg}$ and initial speed $v = 10 \text{ m/s}$ . $K_i = \frac{1}{2}mv^2 = \frac{1}{2}(10)(10)^2 = 500 \text{ J}$ (Refer to NCERT Class 11, Section 5.4, Eq. 5.5).
Calculate Work Done by Variable Force ( $W$ ): The force is variable ( $F = 0.1x$ ) and retarding (opposes motion). The work done by a variable force is calculated by integration. $W = \int_{x_i}^{x_f} F(x) dx$ Since it is a retarding force, we take it as negative or subtract the work magnitude from initial energy. $W = \int_{20}^{30} (-0.1x) dx = -0.1 \left[ \frac{x^2}{2} \right]_{20}^{30}$ $W = -0.05 (30^2 - 20^2) = -0.05 (900 - 400)$ $W = -0.05 (500) = -25 \text{ J}$ (Refer to NCERT Class 11, Section 5.5, Eq. 5.7 regarding work done by variable force).
Apply Work-Energy Theorem: The change in kinetic energy is equal to the work done. $K_f - K_i = W$ $K_f = K_i + W = 500 + (-25) = 475 \text{ J}$ (Refer to NCERT Class 11, Section 5.6 for the Work-Energy Theorem).

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