Answer
the gravitational potential energy is
$$
m g y=m g L(1-\cos \theta)
$$
When $\theta=0^{\circ}$ (the string at its lowest point) we are told that its speed is $8.0 \mathrm{m} / \mathrm{s} ;$ its kinetic $\text { energy there is therefore } 64 \mathrm{J} \text { (using Eq. } 7-1) .$ At $\theta=60^{\circ}$ its mechanical energy is
$$
E_{\text {mech }}=\frac{1}{2} m v^{2}+m g L(1-\cos \theta)
$$
Energy conservation (since there is no friction) requires that this be equal to 64 $\mathrm{J}$ .
Solving for the speed, we find $v=5.0 \mathrm{m} / \mathrm{s}$ .
Work Step by Step
the gravitational potential energy is
$$
m g y=m g L(1-\cos \theta)
$$
When $\theta=0^{\circ}$ (the string at its lowest point) we are told that its speed is $8.0 \mathrm{m} / \mathrm{s} ;$ its kinetic $\text { energy there is therefore } 64 \mathrm{J} \text { (using Eq. } 7-1) .$ At $\theta=60^{\circ}$ its mechanical energy is
$$
E_{\text {mech }}=\frac{1}{2} m v^{2}+m g L(1-\cos \theta)
$$
Energy conservation (since there is no friction) requires that this be equal to 64 $\mathrm{J}$ .
Solving for the speed, we find $v=5.0 \mathrm{m} / \mathrm{s}$ .
