Question

A worker is raising himself and the crate on which he stands with an acceleration of $5{\text{ms}}^{-2}$ by a massless rope and pulley arrangement. Mass of the worker is $100\text{kg}$ and that of the crate is $50\text{kg}$. If $T$ is the tension in the rope and $F$ be the force of contact between the worker and the floor and if $g=10{\text{ms}}^{-2}$, then

• A. $T=2250\text{N},$ $F=1125\text{N}$
• B. $T=1125\text{N},$ $F=2250\text{N}$
• C. $T=1125\text{N},$ $F=375\text{N}$
• D. $T=1125\text{N},$ $F=750\text{N}$

Answer 1

The correct option is C $T=1125\text{N},$ $F=375\text{N}$

$M$ is the mass of worker and $m$ being the mass of crate.

F.B.D of system:


Considering the worker and crater as system and applying newton's $2^{nd}$ law for system.

$T+T-(M+m)g=(M+m)a$

Here, $M=100\text{kg};m=50\text{kg}$

$\Rightarrow 2T-(100+50)10=150\times 5$

$\Rightarrow 2T=1500+750$

$\therefore T=\frac{2250}{2}=1125\text{N}$

Now from the FBD of worker,


Applying Newton's $2^{nd}$ law,

$\Rightarrow T+F-Mg=Ma$

$\Rightarrow 1125+F-\left(100\right)10=100\times 5$

$\Rightarrow 1125+F-1000=500$

$\therefore$ $F=500-125=375\text{N}$

$\begin{array}{c}Ans.C\end{array}$

$\begin{array}{c}\text{Why this question}?\\ \text{Always try to apply newton's second}\\ \text{law for system of bodies, when a lot of forces are}\\ \text{coming in picture. It will eliminate the}\\ \text{internal forces in dynamics equation and}\\ \text{solution becomes simple.}\end{array}$