The bob of a pendulum is released from a hori | vedclass

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(C) Length of the pendulum,$l = 1.5 \; m$.Mass of the bob $= m$.Energy dissipated $= 5 \%$.At the horizontal position,the potential energy is $E_P = m

Vedclass · Accepted Answer

$5.28$

Vedclass · Answer

(C) Length of the pendulum,$l = 1.5 \; m$.Mass of the bob $= m$.Energy dissipated $= 5 \%$.At the horizontal position,the potential energy is $E_P = mgl$ and kinetic energy is $E_K = 0$. Total initial energy $E_i = mgl$.At the lowermost point,the potential energy is $0$ and kinetic energy is $E_K = \frac{1}{2}mv^2$. Total final energy $E_f = \frac{1}{2}mv^2$.Since $5 \%$ of the initial energy is dissipated,the final energy is $95 \%$ of the initial energy:$E_f = 0.95 	imes E_i$$\frac{1}{2}mv^2 = 0.95 	imes mgl$$v^2 = 2 	imes 0.95 	imes g 	imes l$$v = \sqrt{2 	imes 0.95 	imes 9.8 	imes 1.5}$$v = \sqrt{27.93} \approx 5.28 \; m/s$.

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5 \; m$,what is the speed (in $m/s$) with which the bob arrives at the lowermost point,given that it dissipated $5 \%$ of its initial energy against air resistance?

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