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(D) The kinetic energy $(K)$ of a body is given by the formula $K = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the magnitude of the velocity.F

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$6$

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(D) The kinetic energy $(K)$ of a body is given by the formula $K = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the magnitude of the velocity.First, calculate the magnitude of the velocity vector $\vec{v} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) 	ext{ m/s}$.$v = |\vec{v}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} 	ext{ m/s}$.Thus, $v^2 = 29 	ext{ m}^2/	ext{s}^2$.Given $K = 87 	ext{ J}$, substitute the values into the kinetic energy formula:$87 = \frac{1}{2} 	imes m 	imes 29$.$87 = 14.5 	imes m$.$m = \frac{87}{14.5} = 6 	ext{ kg}$.Therefore, the mass of the body is $6 	ext{ kg}$.

If the kinetic energy of a body moving with a velocity of $(2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \text{ m/s}$ is $87 \text{ J}$, then the mass of the body is: (in $\text{ kg}$)

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