91Ó°ÊÓ

1. A mass of pendulums, $\text{M=2.0kg}$
2. Length of string,$L=40m$
3. Velocity at its lowest point is,role="math" localid="1663122028568" $\text{V=80m/s}$

Using the energy conservation law, find the velocity at the lowest point. So, according to the conditions in the problem, find the velocity at the lowest point and the greatest angle of string also. According to the law of energy conservation, energy can neither be created nor destroyed.

Formula:

$$\begin{gathered} Energy=PE+KE \\ =\frac{1}{2}m{v}^2+mgh \end{gathered}$$

where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity and h is height.

From the figure,

$\begin{array}{l}h=L-L\left(\cos \mathrm{θ}\right)\\ h=L\left(1-\left(\cos \mathrm{θ}\right)\right)\end{array}$

The total energy of the system is,

$$\begin{gathered} E_{total}=K{E}_{lowest} \\ =P{E}_{top} \\ ={\left(KE+PE\right)}_{Medieval} \end{gathered}$$

At the top, the object has only potential energy, and at the lowest point, it has only kinetic energy. Therefore,

$$\begin{gathered} E=\frac{1}{2}m{v}^2 \\ E=\frac{1}{2}\left(2\right){\left(8\right)}^2 \\ E=64\text{J} \end{gathered}$$

From the law of conservation of energy, it is clear that,

$$\begin{gathered} E=KE+PE \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mgh \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mg\left(L\left(1-\left(\cos \mathrm{θ}\right)\right)\right) \\ 64\text{J}=\frac{1}{2}m{v}_0^2+mgL-mgL\cos \mathrm{θ} \\ 64\text{J}=\frac{1}{2}\left(2.0\right)v_0^2+\left(2.0\right)\left(9.8\right)\left(4.0\right)-\left(2.0\right)\left(9.8\right)\left(4.0\right)\cos 60 \\ 64\text{J}=v_0^2+78.4-39.2 \\ {v}_0^2=24.8 \\ {v}_0=4.97 \\ =5\text{m/s} \end{gathered}$$

Hence, the speed of pendulum when the string is at to the vertical is$\text{5m/s.}$

At the greatest angle, the velocity of stone will be zero. So, it has only potential energy.

So,

$$\begin{gathered} E=\frac{1}{2}m{v}^2+mgh \\ E=0+mgh \\ E=mgL\left(1-\left(\cos \mathrm{θ}\right)\right) \\ 64=\left(2.0\right)\left(9.8\right)\left(4.0\right)\left(1-\cos \mathrm{θ}\right) \\ \frac{64}{78.4}=\left(1-\cos \mathrm{θ}\right) \\ \cos \mathrm{θ}=1-0.8163 \\ \mathrm{θ}={\cos }^{-1}\left(0.1836\right) \\ \mathrm{θ}=79.4° \end{gathered}$$

Hence, the greatest angle with the vertical that the string will reach during stone’s motion is$79.4°.$

As, it is already calculated,theenergy ofthesystem is$\text{64J}$ , and the law of conservation of energy says that the energy ofthesystem will remain constant. So, the total mechanical energy is$\text{64J}$.

Hence, total mechanical energy of system is$\text{64J}$