Question

Two blocks of masses 4.5 kg and 5.5 kg are connected to the two ends of a light inextensible string passing over a frictionless pulley fixed to a rigid support. At time $t=0$, the blocks are released from rest. The distance travelled by the larger block in a time $t=4$ s is (Acceleration due to gravity $=10\,ms^{-2}$)

• A. 8 m
• B. 4 m
• C. 16 m
• D. 32 m

Hint:

For an Atwood machine: \[ a=\frac{m_2-m_1}{m_1+m_2}g \] and then use standard kinematics equations.

Answer 1

The Correct Option is A

Concept: This is an Atwood machine problem. Acceleration is \[ a=\frac{m_2-m_1}{m_1+m_2}g \]

Step 1: Calculate acceleration.
\[ a=\frac{5.5-4.5}{5.5+4.5}\times10 \] \[ =\frac{1}{10}\times10 \] \[ a=1\,ms^{-2} \]

Step 2: Use equation of motion.
Initially, \[ u=0 \] \[ s=ut+\frac12 at^2 \] \[ s=0+\frac12(1)(4)^2 \] \[ s=8\,m \] Therefore, \[ \boxed{8\,m} \]