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:

${\displaystyle \mathrm{\Delta }{h}_1=s\cdot \sin \theta ,\mathrm{\Delta }{h}_2=-s\cdot \sin \theta }$

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

${\displaystyle \mathrm{\Delta }{E}_p=\mathrm{\Delta }{h}_1{W}_1-\mathrm{\Delta }{h}_2{W}_2}$

By the definition of work:

${\displaystyle F\cdot s=\mathrm{\Delta }{E}_K}$

Then:

${\displaystyle \begin{array}{rl}{\displaystyle F}& {\displaystyle =\frac{\mathrm{\Delta }{E}_K}{s}}\\ {\displaystyle F}& {\displaystyle =\sin \theta \cdot m_1\cdot g-\sin \theta \cdot m_2\cdot g}\\ {\displaystyle F}& {\displaystyle =\sin \theta \cdot g\cdot (m_1-m_2)}\end{array}}$

For the acceleration along the plane, using $a =$ and the total mass $m = m_1 + m_2$, the acceleration is:

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

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

${\displaystyle D=\frac{a{t}^2}{2}}$

Thus:

${\displaystyle \begin{array}{rl}{\displaystyle t^2}& {\displaystyle =\frac{2D}{a}}\\ {\displaystyle t}& {\displaystyle =\sqrt{\frac{2D}{a}}}\end{array}}$

The speed at time $t$ is given by:

${\displaystyle \begin{array}{rl}{\displaystyle v}& {\displaystyle =at}\\ {\displaystyle v}& {\displaystyle =a\cdot \sqrt{\frac{2D}{a}}}\\ {\displaystyle v}& {\displaystyle =\sqrt{2Da}}\\ {\displaystyle v}& {\displaystyle =\sqrt{2D\cdot \sin \theta \cdot g\cdot \frac{m_1-m_2}{m_1+m_2}}}\end{array}}$

For Exercise 2.15, the acceleration is given by:

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

Substituting this into the expression for speed, we get:

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