Question

A block of mass $48\, kg$ kept on a smooth horizontal surface is pulled by a rope of length $4 \,m$ by a horizontal force of $25 \,N$ applied to the other end. If the linear density of the rope is $0.5 \,kg \,m ^{-3}$, the force acting on the block is

• A. 24 N
• B. 25 N
• C. 12 N
• D. 13 N

Answer 1

Answer: (A)

According to the question,

Given, linear density of rope, $\lambda=0.5\, kgm ^{-3}$
hence, mass of rope, $m=\lambda \cdot l$
$=0.5 \times 4=2\, kg$
Therefore, the total mass of system,
$M=48+2=50 \,kg$

$\therefore $ Acceleration, $a=\frac{F}{M}=\frac{25}{50}=\frac{1}{2} ms ^{-2}$
Hence, the force acting on the block is,
$F=48 \times \frac{1}{2}=24\, N$