Question

A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs $1,000$N exerts a force of $450$ N on the chair downwards while pulling the rope on the other side. If the chair weighs $250$N, then the acceleration of the chair is?

• A. $0.45$ $m/s^2$
• B. $0$
• C. $2m/s^2$
• D. $9/25m/s^2$

Answer 1

Answer: (C)

$2m/s^2$

Given weight of the man= 1000 N,thus mass= 100 kg,

weight of the chair=250 N= 25 kg (taking acceleration due to gravity =10 m/$s^2$ and downward force exerted by the chair =450 N

Therefore net force acting on the chair $F_{net}$=$2T-mg-Mg=2T-1250$___(1)

Also we know force= mass$\times$ acceleration

$F_{net}$={mass of chair +mass of man}$\times$acceleration(a)

=>$2T-1250=125a$

=>T$=\frac{125a+1250}{2}$ --------(2)

Now net force on the chair $F_{chair}$=$T-mg-\text{normal force}$

$=T-250-450$

$=T-700$---------(3)

Also $F_{chair}$=mass$\times$acceleration

$T-700=25a$

Now putting the value of T from equation 2 we get

$=>\frac{125a+1250}{2}-700=25a125a=2(25a+700)-1250a=\frac{150}{75}=2m/s^2$