A particle of mass m kg moves along the X-ax | vedclass

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(B) According to the work-energy theorem, the total work done by all forces is equal to the change in kinetic energy of the particle.Work done $W = \D

Vedclass · Accepted Answer

$\frac{m k^2}{2} d^{2 \beta}$

Vedclass · Answer

(B) According to the work-energy theorem, the total work done by all forces is equal to the change in kinetic energy of the particle.Work done $W = \Delta KE = KE_f - KE_i$.The velocity of the particle is given by $v = k x^\beta$.At $x = 0$, the initial velocity $v_i = k(0)^\beta = 0$.At $x = d$, the final velocity $v_f = k d^\beta$.The initial kinetic energy $KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0$.The final kinetic energy $KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (k d^\beta)^2 = \frac{1}{2} m k^2 d^{2\beta}$.Therefore, the total work done $W = \frac{1}{2} m k^2 d^{2\beta} - 0 = \frac{m k^2}{2} d^{2\beta}$.

$A$ particle of mass $m \ kg$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to

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