Question
Two blocks of masses kg and kg are connected by a light inextensible string passing over a frictionless pulley (Atwood machine). Find the acceleration of the system and the tension in the string. Take m/s.
The diagram is the standard Atwood machine: hanging on one side, on the other, string over a single pulley at the top. Let’s solve it the FBD way — that’s what JEE expects.
Solution — Step by Step
Since , the heavier block goes down and the lighter one goes up. Let acceleration magnitude be . Both blocks have the same because the string is inextensible.
For (taking downward positive, since it accelerates down):
For (taking upward positive, since it accelerates up):
From N.
Final answer: m/s, N.
Why This Works
The constraint that both blocks have the same acceleration magnitude comes directly from the string being inextensible — if moves down by , must move up by in the same time. Differentiate twice and the accelerations match.
The tension is the same throughout the string because the pulley is frictionless and the string is light (massless). If either of those changes, tensions on the two sides differ.
For masses on a frictionless light pulley:
Alternative Method
Treat the system as one unit. Net unbalanced force N. Total mass kg. So m/s. Faster, but only works for finding — for , you still need an FBD.
Students often write or thinking the string just “holds up” the block. Wrong — the system is accelerating, so . Use depending on which block you’re looking at.
Common Mistake
The biggest trap on diagram-based Newton’s laws questions: assigning the wrong direction of acceleration in the FBD. If you guess wrong, you’ll get a negative (which is fine — just means you guessed the direction wrong) but students panic and redo the whole problem instead of trusting the math. Sign tells direction; magnitude is what you report.