A sphere of mass m is attached to a spring of | vedclass

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(A) Let $x$ be the maximum extension of the spring.Since the sphere starts from rest and comes to rest momentarily at the maximum extension,the total

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$\frac{2 m g \sin 	heta}{k}$

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(A) Let $x$ be the maximum extension of the spring.Since the sphere starts from rest and comes to rest momentarily at the maximum extension,the total work done by all forces is zero.The forces acting on the sphere are gravity,the spring force,and the normal force.The work done by the normal force is zero as it is perpendicular to the displacement.The work done by gravity is $W_g = m g x \sin 	heta$.The work done by the spring is $W_s = -\frac{1}{2} k x^2$.By the work-energy theorem,the change in kinetic energy is equal to the total work done:$0 - 0 = W_g + W_s$$m g x \sin 	heta - \frac{1}{2} k x^2 = 0$$m g \sin 	heta = \frac{1}{2} k x$$x = \frac{2 m g \sin 	heta}{k}$

$A$ sphere of mass $m$ is attached to a spring of spring constant $k$ and is held in an unstretched position over an inclined plane as shown in the figure. After letting the sphere go,find the maximum length by which the spring extends,given the sphere only rolls.

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