Question
A block of mass kg sits on a frictionless table. A second block of mass kg hangs over a pulley at the edge, connected by a light string. Find the acceleration of the system and the tension in the string. Take m/s.
Solution — Step by Step
For on the table: tension pulls horizontally; weight and normal cancel vertically. For : weight pulls down, tension pulls up. Both blocks have the same magnitude of acceleration because the string is inextensible.
For : . For : . Notice that we wrote the net force in the direction of motion for each block — that’s the key.
Final answer: m/s, N.
Why This Works
The trick is treating each block as a separate system but using the constraint that they share the same acceleration magnitude. Many students try to use , assuming the hanging block is in equilibrium — but it isn’t, it’s accelerating downward.
The system equation is worth memorising for Atwood-type pulley problems. It tells you the net driving force divided by the total inertia.
Alternative Method
Treat both blocks as one system with total mass and net external force (only gravity on the hanging block contributes, since tension is internal). Then directly. Use this shortcut once you trust it — but in JEE Advanced, write the FBD anyway to avoid sign errors.
The most common slip: writing for the hanging block. That’s only true in equilibrium. Since the block accelerates, . Always check: is the system actually in equilibrium, or is it accelerating?
Common Mistake
Students sometimes assume the tension on differs from the tension on . For a massless, inextensible string over a frictionless, massless pulley, tension is uniform throughout. If the pulley has mass or the string has mass, then tensions differ — but those are advanced cases flagged explicitly in the question.