Question

Two blocks of mass 20 kg and 10 kg are connected by a massless string as shown in the figure. How much time will the 20 kg block take to travel 0.6 m distance when released from rest?
[Take gravitational acceleration as $10m/s^2]$

• A. 0.2 s
• B. 0.4 s
• C. 1 s
• D. 0.6 s

Answer 1

The correct option is D 0.6 s

Let the tension in the string be T and acceleration of the blocks be a as shown in the figure.
For block of mass $m_1=10kg:$
$T-m_{1}g=m_{1}a$
$\Rightarrow T-10g=10a...\left(i\right)$
For block of mass $m_2=20kg$
$m_{2}g-T=m_{2}a$
$\Rightarrow 20g-T=20a...\left(ii\right)$
Adding (i) and (ii), we get:
30a = 10g
$\Rightarrow a=\frac{1}{3}g=\frac{10}{3}m/s^2(\because g=10m/s^2)$
For 20 kg block:
Initially velocity, u = 0
Displacement, s = 0.6 m
Acceleration, $a=\frac{10}{3}m/s^2$
Using second equation of motion, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=0\times t\frac{1}{2}a{t}^2$
$\Rightarrow t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 0.6}{\left(\frac{10}{3}\right)}}$
$\Rightarrow t=0.6s$
Hence, the correct answer is option (4).