Potential energy of a body of mass 1 kg free | vedclass

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Question

(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.Given,$E = 2 \ J$

Vedclass · Accepted Answer

$\sqrt{5} \ ms^{-1}$

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(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.Given,$E = 2 \ J$ and $m = 1 \ kg$.To find the maximum speed,we need the maximum kinetic energy,which occurs when the potential energy is at its minimum.$U(x) = \frac{x^2}{2} - x$.To find the minimum potential energy,we set $\frac{dU}{dx} = 0$:$\frac{d}{dx}(\frac{x^2}{2} - x) = x - 1 = 0 \implies x = 1 \ m$.$U_{\min} = U(1) = \frac{1^2}{2} - 1 = -0.5 \ J$.Since $E = U + KE$,we have $KE_{\max} = E - U_{\min} = 2 - (-0.5) = 2.5 \ J$.Using $KE = \frac{1}{2}mv^2$:$2.5 = \frac{1}{2}(1)v^2 \implies v^2 = 5 \implies v = \sqrt{5} \ ms^{-1}$.

Potential energy of a body of mass $1 \ kg$ free to move along $X$-axis is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the body is $2 \ J$,then the maximum speed of the body is (Assume only conservative force acts on the body).

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