Feynman’s Lectures Exercises 2.14 and 2.15

If the weights W1W_1 and W2W_2 move the distance ss along the plane to the left, their corresponding difference in height are:

Δh1=ssinθ,Δh2=ssinθ

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

ΔEp=Δh1W1Δh2W2

By the definition of work:

Fs=ΔEp

Then:

F=ΔEps=sinθm1gsinθm2g=sinθg(m1m2)

For the acceleration along the plane, using a=Fma = and the total mass m=m1+m2m = m_1 + m_2, the acceleration is:

a=sinθgm1m2m1+m2

For the motion over distance DD, starting from rest:

D=at22

Thus:

t2=2Dat=2Da

The speed at time tt is given by:

v=at=a2Da=2Da=2Dsinθgm1m2m1+m2

For Exercise 2.15, the acceleration is given by:

a=12g(sinθsinϕ)

Substituting this into the expression for speed, we get:

v=gD(sinθsinϕ)