Question

In the given picture both pulley and string are massless. Pulley is frictionless. Find the acceleration of mass 2m

• A. g
• B. 2g
• C.
• D.

Answer 1

The correct option is C

Constraint relation:-

If 2m moves x down then m will move x up. Pretty obvious right

And hence if 2m has acceleration a downwards then m will have acceleration a upwards. At times if might become not that obvious. So we have a technique.

(1) Take reference point (Here the pulley)

(2) (1)Measure the length of the string by assigning variables.

Here: - Let length of string connected to block of mass 2m be $x_1$

Total length of the string is l = $x_1$ + $x_2$

Assume acceleration of the blocks between which you want constraint relation.

$\frac{dl}{dt}=\frac{d{x}_1}{dt}+\frac{d{x}_2}{dt}\Rightarrow 0=-V_1+V_2$

$\frac{d{x}_1}{dt}=-V_1$ as $x_1$ is being decreased by upward acceleration $a_1$

$\frac{d{x}_2}{dt}=V_2$ as $x_2$ is being increased by downward acceleration $a_2$

$(\frac{dl}{dt}=0aslisconstant)$ Differentiating once again

0 = -$a_1+a_2$

$a_1$ = $+a_2$

So if 2m block goes down with an acceleration a then block m goes up with same acceleration a

Draw free body drawing write $e{q}^n$

T -mg = ma ____(i) 2mg - T = 2ma ___(ii)

Adding (i) and (ii)

mg = 3ma

$a=\frac{mg}{3m}=\frac{g}{3}$