A pendulum bob has a speed 4 ,m/s at its lowe | vedclass

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(C) Let the length of the pendulum be $L = 1 \,m$ and the initial speed at the lowest point be $v_0 = 4 \,m/s$.At the lowest point, the potential ener

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$\sqrt{6} \,m/s$

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(C) Let the length of the pendulum be $L = 1 \,m$ and the initial speed at the lowest point be $v_0 = 4 \,m/s$.At the lowest point, the potential energy is taken as $0$. The total energy is $E = \frac{1}{2}mv_0^2$.When the string makes an angle $	heta = 60^{\circ}$ with the vertical, the height of the bob is $h = L(1 - \cos 	heta)$.Substituting the values, $h = 1(1 - \cos 60^{\circ}) = 1(1 - 0.5) = 0.5 \,m$.By the law of conservation of energy, the total energy at the lowest point equals the total energy at the angle $	heta$:$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$$v^2 = v_0^2 - 2gh$$v^2 = (4)^2 - 2(10)(0.5)$$v^2 = 16 - 10 = 6$$v = \sqrt{6} \,m/s$.

$A$ pendulum bob has a speed $4 \,m/s$ at its lowest position. The pendulum is $1 \,m$ long. When the length of the string makes an angle of $60^{\circ}$ with the vertical, the speed of the bob at that position is (acceleration due to gravity, $g=10 \,m/s^2, \cos 60^{\circ}=0.5$).

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