Question

A rope of mass $m$ is attached to a block of mass $M$ lying on a horizontal surface. The block is pulled along the surface by applying a force $F$ on the free end of the rope. If $\mu$ is the coefficient of friction between block and the surface, the force exerted by the rope on the block is

• A. $\frac{m}{(M+m)}(F-\mu mg)$
• B. $\frac{M}{(M+m)}(F-\mu Mg)$
• C. $\frac{m}{M}(F+\mu mg)$
• D. $\frac{M}{(m+M)}(F+\mu Mg)$

Answer 1

The correct option is D $\frac{M}{(m+M)}(F+\mu Mg)$

The free-body diagrams of the block and the rope are shown in Fig.

For block : $T-f=Ma$ where $f=\mu Mg$
$\therefore T-\mu Mg=Ma.........\left(1\right)$
For rope:
$F-T=ma........\left(2\right)$
Dividing (1) by (2) and simplifying,
$\frac{T-\mu Mg}{F-T}=\frac{M}{m}$
$\Rightarrow T-\mu Mg=\frac{M}{m}F-T\frac{M}{m}$
$\Rightarrow T[1+\frac{M}{m}]=\frac{M}{m}[F+\mu mg]$
$\Rightarrow T=\frac{M}{(M+m)}[F+\mu mg]$
we find that the correct choice is (d).