Position of a 2 ,kg mass moving along the X-a | vedclass

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(D) The position of the particle is given by $x = 2 \cos (2t)$.Taking the derivative with respect to time $t$,we get the velocity $v$:$v = \frac{dx}{d

Vedclass · Accepted Answer

$16$

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(D) The position of the particle is given by $x = 2 \cos (2t)$.Taking the derivative with respect to time $t$,we get the velocity $v$:$v = \frac{dx}{dt} = \frac{d}{dt} [2 \cos (2t)] = -2 \sin (2t) 	imes 2 = -4 \sin (2t) \,m/s$.The maximum velocity $v_{\max}$ occurs when $|\sin (2t)| = 1$,so $|v_{\max}| = 4 \,m/s$.The maximum kinetic energy $K_{\max}$ is given by the formula $K_{\max} = \frac{1}{2} m (v_{\max})^2$.Substituting the given mass $m = 2 \,kg$ and $v_{\max} = 4 \,m/s$:$K_{\max} = \frac{1}{2} 	imes 2 \,kg 	imes (4 \,m/s)^2 = 16 \,J$.

Position of a $2 \,kg$ mass moving along the $X$-axis is given by $x=2 \cos (2 t) \,m$. Then the maximum kinetic energy of the mass in joules is

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