Statement (I): The slope of the kinetic energ | vedclass

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(C) Analysis of Statement $(I)$: The kinetic energy $K = \frac{1}{2}mv^2$. The slope of the $K-x$ curve is $\frac{dK}{dx} = \frac{d}{dx}(\frac{1}{2}mv

Vedclass · Accepted Answer

Statement $I$ is true, but statements $II$ and $III$ are false

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(C) Analysis of Statement $(I)$: The kinetic energy $K = \frac{1}{2}mv^2$. The slope of the $K-x$ curve is $\frac{dK}{dx} = \frac{d}{dx}(\frac{1}{2}mv^2) = mv \frac{dv}{dx} = m \cdot a$. Since $m$ is constant, the slope is directly proportional to acceleration $a$. Thus, Statement $(I)$ is true.Analysis of Statement $(II)$: Initial velocity $u = 30 \ m/s$, height $h = 15 \ m$, $g = 10 \ m/s^2$. Velocity just before hitting the ground $v^2 = u^2 + 2gh = 30^2 + 2(10)(15) = 900 + 300 = 1200 \ (m/s)^2$. Kinetic energy before impact $E_i = \frac{1}{2}m(1200)$. After impact, it rises to $15 \ m$, so velocity after impact $v' = \sqrt{2gh} = \sqrt{2(10)(15)} = \sqrt{300} \ m/s$. Kinetic energy after impact $E_f = \frac{1}{2}m(300)$. Loss in energy $\Delta E = E_i - E_f = \frac{1}{2}m(1200 - 300) = \frac{1}{2}m(900)$. Percentage loss $= (\frac{\Delta E}{E_i}) 	imes 100 = (\frac{900}{1200}) 	imes 100 = 75 \%$. Thus, Statement $(II)$ is false.Analysis of Statement $(III)$: From $v^2 = u^2 + 2as$, where $u=0$ and $a = F/m$, we get $v^2 = 2(F/m)s$. Thus $v = \sqrt{2Fs/m}$. Velocity is inversely proportional to $\sqrt{m}$, not directly proportional to $m$. Thus, Statement $(III)$ is false.

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