Question
A block of mass rests on a frictionless table. It is connected by a light, inextensible string passing over a frictionless pulley to a hanging block of mass . Find the acceleration of the system and the tension in the string. Take .
Solution — Step by Step
For on the table, the only horizontal force is the tension pulling it forward. For hanging, gravity pulls it down and tension pulls it up. The string ensures both blocks have the same magnitude of acceleration .
For : . For : .
We treat each block separately because they have different net forces, but the acceleration is shared.
Add the two equations to eliminate :
Final answer: , .
Why This Works
Newton’s second law applies to each body individually. The string constraint links them — same magnitude of acceleration because the string doesn’t stretch. The pulley merely changes the direction of tension; with a massless, frictionless pulley, the tension is the same on both sides.
This Atwood-style setup is one of the most common JEE/NEET problems because it tests free-body discipline. Once we get the FBD right, the algebra is mechanical.
Alternative Method
Treat the system as one body with total mass . The only external force driving motion is (the weight of the hanging block; weight of is balanced by the normal force).
This shortcut works only when both masses move with the same speed. To find , we must still go back to one block’s FBD.
A common mistake is to write , treating the hanging block as if it were in equilibrium. It isn’t — it’s accelerating downward, so the net force on it cannot be zero. Always check: if the string accelerates, .
Common Mistake
Students often plug in the middle of the problem after starting with . Pick one and stick with it. JEE Main usually accepts unless the problem says otherwise.