Work Done by the Gravitational Force — Question 8

Question 8

Opposite height changes make the net gravity work a mass difference

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. Masses of 3 kg and 5 kg hang on an ideal massless cord over a fixed frictionless massless pulley. Released from rest, the 5 kg mass descends 0.50 m and the 3 kg mass rises equally.

1. Find gravity work on each mass and its sum.

2. Find the common speed after this motion.

3. Explain why tension can be omitted from the summed work equation but not from either separate equation.

Question 8 - Solution

Opposite height changes make the net gravity work a mass difference

(a) Assign the two height-change signs.

The governing relation is $$\boxed{W_{g,\mathrm{total}}=(m_2-m_1)gd=\tfrac12(m_1+m_2)v^2}$$ The 5 kg mass receives $$+5(10)(.50)=25\ \mathrm J.$$ The 3 kg mass receives $$-3(10)(.50)=-15\ \mathrm J.$$ Net gravity work is 10 J.

(b) Sum the two kinetic energies.

The cord imposes equal speeds, so $$10=\tfrac12(3+5)v^2.$$ Thus $$v=\sqrt{2.5}\ \mathrm{m/s}.$$

(c) Use the ideal cord constraint.

Tension is equal at both ends; it does $-T(.50)$ on the descending mass and $+T(.50)$ on the ascending mass. These works cancel only in the sum. Each separate mass still has tension work.