Feynman’s Lectures Exercises 2.14 and 2.15

If the weights $W_1$ and $W_2$ move the distance $s$ along the plane to the left, their corresponding difference in height are:

$$\Delta h_1 = s \cdot \sin{\theta} , \quad \Delta h_2 = - s \cdot \sin{\theta}$$

Then, the difference in potential energy of the system is:

$$\Delta E_p = \Delta h_1 W_1 - \Delta h_2 W_2$$

By the definition of work:

$$F \cdot s = \Delta E_p$$

Then:

$$\begin{aligned} F &= \frac{\Delta E_p}{s} \\[10pt] &= \sin{\theta} \cdot m_1 \cdot g - \sin{\theta} \cdot m_2 \cdot g \\[10pt] &= \sin{\theta} \cdot g \cdot (m_1 - m_2) \end{aligned}$$

For the acceleration along the plane, using $a = \frac{F}{m}$ and the total mass $m = m_1 + m_2$, the acceleration is:

$$a = \sin{\theta} \cdot g \cdot \frac{m_1 - m_2}{m_1+m_2}$$

For the motion over distance $D$, starting from rest:

$$D = \frac{at^2}{2}$$

Thus:

$$\begin{aligned} t^2 &= \frac{2D}{a} \\[10pt] t &= \sqrt{\frac{2D}{a}} \end{aligned}$$

The speed at time $t$ is given by:

$$\begin{aligned} v &= at \\[0pt] &= a \cdot \sqrt{\frac{2D}{a}} \\[10pt] &= \sqrt{2Da} \\[2pt] &= \sqrt {2D \cdot \sin{\theta} \cdot g \cdot \frac{m_1 - m_2}{m_1 + m_2}} \end{aligned}$$

For Exercise 2.15, the acceleration is given by:

$$a = \frac{1}{2} \cdot g \cdot (\sin{\theta} - \sin{\phi})$$

Substituting this into the expression for speed, we get:

$$v = \sqrt{g \cdot D \cdot (\sin{\theta} - \sin{\phi})}$$