Question

Two objects of masses 1 kg and 3 kg are connected by a massless string passing over a fixed smooth pulley The acceleration of heavier mass with respect to the lighter mass is

Answer 1

Acceleration of the system

m1 = 1 kg

m2 = 3 kg

$a \: = \frac{(m2 - m1)}{(m1 + m2)} g \\ \\ a \: = \frac{(3 -2) }{(3 + 2)}10 \\ \\ a \: = 5m \: {s}^{ - 2}$

Hence acceleration of heavier mass = 5 m/s^-2

Answer 2

Given :

Two object of masses 1 kg and 3 kg

To find:

Acceleration of the heavier mass with respect to the lighter mass.

Solution:

Since, it is given that the two masses are connected by string, Let the heavier mass of 3 kg be $m_{1}$ and that the lighter mass of 1 kg be $m_{2}$.

Step 1

If we consider the FBD ( free body diagram ) of the heavier mass.

$m_{1} g-T=m_{1} a$  ; or

$T =m_{1}g - m_{1}a$

Similarly,

FBD of the lighter mass.

$T-m_{2}g=m_{2}a$  ; or

$T =m_{2}g+m_{2}a$

Equating the only tension that will produce in the string,

$m_{2}g+m_{2}a=m_{1}g-m_{1}a$

$a=\frac{(m_{1} -m_{2} )g}{(m_{1} +m_{2} )}$

Substituting the values, we get

$a =\frac{(3-1)10}{3+1} \\a = 5m/s^{2}$

Final answer:

Since, $m_{1}$ mass is heavier than $m_{2}$ hence, $m_{1}$ will go downwards with acceleration of $5 m/s^{2}$ and pull the lighter mass $m_{2}$ upwards.