Question:
Two masses $ m_{1}=5\,kg $ and $ m_{2}=4.8\,kg $ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move ? $ (g=9.8 \, m/s^{2}) $
Two masses $ m_{1}=5\,kg $ and $ m_{2}=4.8\,kg $ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move ? $ (g=9.8 \, m/s^{2}) $
Updated On: Jul 5, 2022
- $ 0.2\,m/s^{2} $
- $ 9.8\,m/s^{2} $
- $ 5\,m/s^{2} $
- $ 4.8\,m/s^{2} $
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The Correct Option is A
Solution and Explanation
On releasing, the motion of the system will be according to figure.
$m_{1} g-T=m_{1} a \ldots$ (i)
and $T-m_{2} g=m_{2} a \ldots$ (ii)
On solving; $a=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g ....$(iii)
Here,
$m_{1}=5 \,kg , m_{2}=4.8 \,kg , g=9.8\, m / s ^{2}$
$ \therefore a=\left(\frac{5-4.8}{5+4.8}\right) \times 9.8$
$=\frac{0.2}{9.8} \times 9.8=0.2\, m / s ^{2}$
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.