A block of mass 2 ~kg is initially at rest on | vedclass

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(C) The work done by the force on the block is equal to the change in its kinetic energy.Since the block starts from rest at $x = 0$,the kinetic energ

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$18$

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(C) The work done by the force on the block is equal to the change in its kinetic energy.Since the block starts from rest at $x = 0$,the kinetic energy $KE$ at any position $x$ is given by the work done by the force $F$ from $x = 0$ to $x$.$KE = \int_{0}^{x} F dx = \int_{0}^{x} (9 - x^2) dx = 9x - \frac{x^3}{3}$.To find the maximum kinetic energy,we find where the force becomes zero:$F = 9 - x^2 = 0 \implies x^2 = 9 \implies x = 3 ~m$.At $x = 3 ~m$,the kinetic energy is:$KE_{max} = \int_{0}^{3} (9 - x^2) dx = [9x - \frac{x^3}{3}]_{0}^{3} = (9(3) - \frac{3^3}{3}) - 0 = 27 - 9 = 18 ~J$.

$A$ block of mass $2 ~kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} ~N$ acts on it,starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 ~m$ in joules is:

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