Two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley as shown in figure $m_{1} = 3 k g$ and $m_{2} = 6 k g$ . The system is released from rest (a) Find the distance traveled by the first block the first two seconds. $g = 10 m / s^{2}$

Open in App

Solution

$Step 1: Drawing FBD$ $[Ref. Fig.]$
$T =$ Tension in string

$Step 2: Applying Newton's Second Law$
On block $2$ (Taking downward positive)
$\sum F_{y} = m a$
$- T + m_{2} g = m_{2} a$ $. . . . (1)$

On block $1$ (Taking upward positive)
$\sum F_{y} = m a$
$T - m_{1} g = m_{1} a$ $. . . . (2)$

Adding Equations $(1)$ and $(2)$ , we get
$m_{2} g - m_{1} g = m_{1} a + m_{2} a$

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

$\Rightarrow$ $a = \frac{(6 - 3) \times 10}{6 + 3} m / s^{2} = \frac{10}{3} m / s^{2}$

$Step 3: Applying equations of motion$
Since acceleration is constant, therefore applying equation of motion for block $1$
(Taking upward positive)
$S = u t + \frac{1}{2} a t^{2}$

$u = 0; t = 2 s$

$S = \frac{1}{2} \times \frac{10}{3} \times (2)^{2} m$

$= 6.66 m$

Hence, the distance traveled by the first block is $6.66 m$ in the first two seconds.

Suggest Corrections

Similar questions

Q. In a simple Atwood machine, two unequal masses

m_{1}

and

m_{2}

are connected by a string going over a clamped light smooth pulley. In a typical arrangement shown in figure,

m_{1} = 300 g

and

m_{2} = 600 g

. The system is released from the rest.Find the distance traveled by the first block in the first two seconds.

In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement (Figure) $m_{1} = 300$ g and $m_{2} = 600$ g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley.