One end of a spring of spring constant 80 Nm | vedclass

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(D) Given: Spring constant $k = 80 \ Nm^{-1}$,unstretched length $l_0 = 0.3 \ m$,mass of ring $m = 0.3 \ kg$,height of rod $h = 0.4 \ m$.Initial state

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) Given: Spring constant $k = 80 \ Nm^{-1}$,unstretched length $l_0 = 0.3 \ m$,mass of ring $m = 0.3 \ kg$,height of rod $h = 0.4 \ m$.Initial state: The spring makes an angle $	heta = 60^{\circ}$ with the vertical. The length of the spring is $l_1 = h / \cos(60^{\circ}) = 0.4 / 0.5 = 0.8 \ m$.The extension in the spring is $x_1 = l_1 - l_0 = 0.8 - 0.3 = 0.5 \ m$.Final state: The spring is vertical. The length of the spring is $l_2 = h = 0.4 \ m$.The extension in the spring is $x_2 = l_2 - l_0 = 0.4 - 0.3 = 0.1 \ m$.Using the law of conservation of mechanical energy: $U_i + K_i = U_f + K_f$.Initial energy $U_i = \frac{1}{2} k x_1^2 = \frac{1}{2} 	imes 80 	imes (0.5)^2 = 40 	imes 0.25 = 10 \ J$.Final energy $U_f = \frac{1}{2} k x_2^2 = \frac{1}{2} 	imes 80 	imes (0.1)^2 = 40 	imes 0.01 = 0.4 \ J$.Since the system is released from rest,$K_i = 0$. Let the final speed be $v$.$10 + 0 = 0.4 + \frac{1}{2} m v^2$.$9.6 = \frac{1}{2} 	imes 0.3 	imes v^2$.$v^2 = (9.6 	imes 2) / 0.3 = 19.2 / 0.3 = 64$.$v = 8 \ ms^{-1}$.

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