Work and Kinetic Energy — Question 6

Question 6

Internal tension cancels only after the subsystem works are added

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. Two carts A (1 kg, in front) and B (2 kg) are connected by an ideal taut horizontal cord on a frictionless track. A constant 12 N force pulls A right. Both start from rest and move 3 m right with the same speed.

1. Find the final common speed from total work.

2. Find the work done by tension on B and the cord tension.

3. Find tension work on A and check its kinetic-energy change.

Question 6 - Solution

Internal tension cancels only after the subsystem works are added

(a) Sum the two cart equations.

The governing relation is $$\boxed{W_{\mathrm{external}}=\Delta K_A+\Delta K_B=\tfrac12(m_A+m_B)v^2}$$ Internal tension works cancel because their displacements are equal and the forces opposite. $$12(3)=\tfrac12(1+2)v^2,$$ so $$v=\sqrt{24}=2\sqrt6\ \mathrm{m/s}.$$

(b) Isolate the trailing cart.

$$\Delta K_B=\tfrac12(2)(24)=24\ \mathrm J.$$ Tension is its only horizontal force, so $3T=24$ and $$T=8\ \mathrm N.$$

(c) Use the correct tension direction on A.

Tension does $-24\ \mathrm J$ on A; the pull does +36 J. Net work is 12 J, equal to $\tfrac12(1)(24)$. The subsystem balances agree with the pair balance.