The variation of force F acting on a body mov | vedclass

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(B) For a body to be in equilibrium,the net force $F$ acting on it must be zero. From the graph,the force $F$ is zero at points $P$ and $Q$.For stable

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$Q$

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(B) For a body to be in equilibrium,the net force $F$ acting on it must be zero. From the graph,the force $F$ is zero at points $P$ and $Q$.For stable equilibrium,the potential energy $U$ must be at a minimum,which implies $\frac{d^2 U}{d x^2} > 0$.Since $F = -\frac{d U}{d x}$,we have $\frac{d F}{d x} = -\frac{d^2 U}{d x^2}$.Therefore,for stable equilibrium,$\frac{d F}{d x} At point $P$,the slope of the $F-x$ curve is positive.At point $Q$,the slope of the $F-x$ curve is negative.Thus,the body is in a stable equilibrium state at point $Q$.

Column-$I$	Column-$II$
$(1)$ Conservative force	$(a)$ Frictional force
$(2)$ Non-conservative force	$(b)$ Gravitational force
	$(c)$ Internal force

The variation of force $F$ acting on a body moving along the $x$-axis with respect to its position $(x)$ is shown in the figure. The body is in a stable equilibrium state at:

Similar Questions

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Conservative force $(a)$ Frictional force

$(2)$ Non-conservative force $(b)$ Gravitational force

$(c)$ Internal force

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The potential energy for a conservative system is given by:
$U = ax^2 - bx$
Where $a$ and $b$ are positive constants. The law of the force governing the system is:

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