Question

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

A.)$0.45m/{s^2}$
B.)$0$
C.)$2m/{s^2}$
D.)$\dfrac{9}{{25}}m/{s^2}$

Answer 1

Hint- This question is based on a pulley problem which is nothing but a special case (i.e. constraint relation) of mechanics. In this problem we will simply use the equation of Newton's second law motion ,which states that net force on a particle or object is equal to the mass times the acceleration (in the direction of net force )of the object .

Complete step-by-step answer:
Mathematically,
F= ma
Where, F- net force on the object.
m-mass of the object
and a- acceleration of the object .

Given

The value of acceleration due to gravity is $10m/{s^2}$
The weight of the man M =1000N, therefore the mass of the men in kg is 100
Weight of the chair m = 250N, mass of the chair in kg is 25kg
The downward force exerted by the chair = 450N

The net force acting on the system of man and chair is

${F_{system}} = 2T - mg - Mg$

Substituting the value of the mass of the chair and the mass of the man and the value of the acceleration due to gravity

${F_{system}} = 2T - 100 \times 10 - 25 \times 10 \\\ {F_{system}} = 2T - 1250..........\left( 1 \right) \\\$

As we know that the force on the chair can be given as

Force = mass x acceleration
Let a be the acceleration of the chair with which it is going downward
Therefore

${F_{system}} = \left( {m + M} \right)a$

Substituting this value in the equation (1)

$(m + M)a = 2T - 1250 \\\ 125a = 2T - 1250 \\\ T = \dfrac{{125a + 1250}}{2}................\left( 2 \right) \\\$

Now net force on the chair is

${F_{chair}} = T - mg - normal\,$

Substituting the value of m and normal force in the above equation

${F_{chair}} = T - 250 - 450 \\\ {F_{chair}} = T - 750..........\left( 3 \right) \\\$

Since

$${F_{chair}} = ma \\\ {F_{chair}} = 25a \\\$$

Therefore equation (3) becomes

$T - 700 = 25a............(4)$

Solving equation (2) and equation (4) we get.

$\Rightarrow \dfrac{{125a + 1250}}{2} - - 700 = 25a \\\ 125a = 2\left( {25a + 700} \right) - 1250 \\\ 75a = 150 \\\ a = \dfrac{{150}}{{75}} = 2m/{s^2} \\\$

The acceleration of the is $2m/{s^2}$

Hence, the correct option is C.

Note– In order to solve these types of questions, you must have a good concept of the laws of motion and keep some points in mind during solving these types of questions. One of those points is Newton’s law of motion is always and only applicable in an inertial frame of reference. It is not applicable in non -inertial frame of reference.