91影视

The given data is listed below as-

• The mass of wagon with gold is, M =300 kg
• The Lone ranger mass is, $m_R=75鈥�\mathrm{kg}$
• The mass of tonto is, $m_T=60鈥�\mathrm{kg}$
• The length of the slope is,l =50.0 m
• Inclination of the slope is $伪=6掳$
• Distance from a cliff is, $d=40鈥�m$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

The law of conservation of mechanical energy is used Since there is no friction,.

From fig,8.99 we have,

$M鈥�gl鈥�sin伪=\frac{1}{1}鈥�M鈥�{v^2}_1$

$v_1=2鈥�gl鈥�sin伪$鈥︹赌︹赌︹赌�.(1)

The total horizontal momentum is conserved since there are no horizontal forces on the wagon.

Their initial horizontal momentum is zero since Ranger and Tonto were vertically falling on wagon.

Let $v_2$be the speed of the wagon after they landed on it.

Therefore,

$M{v}_1=(M+m_R+m_T)v_2$

$v_2=\frac{M}{M+m_R+m_T}{v}_1$鈥︹赌︹赌︹赌�(2)

The time after which the wagon will drop of the cliff after it passed the tree is calculated as below:

$t=\frac{d}{t_2}$

$t=鈥�\frac{d}{\sqrt{gl鈥�sin伪}}\frac{M+m_R+m_T}{M}$

$$\begin{gathered} t=\frac{40}{\sqrt{2脳9.8脳50脳\sin 6掳}}\frac{300+75+60}{300} \\ =5.73s \end{gathered}$$

Now, t >5 s

Therefore, our heroes will manage to save the gold in time.

The total kinetic energy before the landing was,

$K_{before}=K_h+\frac{1}{2}M{v^2}_1$

The kinetic energy after the landing was,

$$\begin{gathered} K_{after}=\frac{1}{2}\left[M+m_R+m_T\right]{v^2}_2 \\ =\frac{1}{2}\frac{M}{M+m_R+m_T}M{v^2}_1 \\ <\frac{1}{2}M{v^2}_1 \end{gathered}$$

Therefore, $K_{after}鈮�K_{before}$

Thus, the total kinetic energy decreases when the two heroes drop into the wagon.