Question

A block of mass 18.5 kg kept on a smooth horizontal surface is pulled by a rope of 3 m length by a horizontal force of 40 N applied to the other end of the rope. If the linear density of the rope is 0.5 kgm\(^-1\) and initially the block is at rest, the time in which the block moves a distance of 9 m is

• A. 3 s
• B. 5 s
• C. 7 s
• D. 9 s

Hint:

For problems involving forces on objects connected by a rope, remember to account for the mass of the rope as well as the object being pulled, and use the total mass to calculate acceleration.

Answer 1

The Correct Option is A

Step 1: The total force applied is \( F = 40 \, \text{N} \). The rope has a linear density \( \mu = 0.5 \, \text{kg/m} \), and the length of the rope is \( L = 3 \, \text{m} \). The total mass of the rope is \( m_{\text{rope}} = \mu L = 0.5 \times 3 = 1.5 \, \text{kg} \). 

Step 2: The total force acting on the system is the sum of the applied force and the force due to the rope's mass. This gives us the total mass \( m_{\text{total}} = 18.5 \, \text{kg} + 1.5 \, \text{kg} = 20 \, \text{kg} \). The acceleration of the system can now be calculated using Newton’s second law: \[ a = \frac{F}{m_{\text{total}}} = \frac{40}{20} = 2 \, \text{m/s}^2 \] 

Step 3: Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \), where \( u = 0 \) (initial velocity) and \( s = 9 \, \text{m} \), we can solve for \( t \): \[ 9 = 0 + \frac{1}{2} \times 2 \times t^2 \] \[ 9 = t^2 \] \[ t = \sqrt{9} = 3 \, \text{s} \] Thus, the time taken for the block to move 9 m is \( 3 \, \text{s} \).