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Question: Two blocks are connected by a massless string through an ideal pulley as shown. A force of \(22\,N\)...

Two blocks are connected by a massless string through an ideal pulley as shown. A force of 22N22\,N is applied on block BB when initially the blocks are at rest. Then acceleration of centre of mass of block AA and block BB, 2s2\,s, after the application of force is (( masses of AA and BB are 4kg4\,kg and 6kg6\,kg respectively and surfaces are smooth ))

A. 1.4ms21.4\,m{s^{ - 2}}
B. 1ms21\,m{s^{ - 2}}
C. 2ms22\,m{s^{ - 2}}
D. None of these.

Explanation

Solution

The point where the whole mass is assumed to be concentrated is known as the centre of mass and it is relative to an object or the system of objects and also the average position of all parts of the system, weighted according to their masses. The tension is nothing but the pulling force in the string and the direction of the string will be away from the load.

Complete step by step answer:
Given: Force exerted on the block BB when the blocks are at rest is =22N = 22\,N
F=22NF = 22\,N
Mass of block A=4kgA = 4\,kg
Mass of block A=6kgA = 6\,kg
We need to find acceleration of centre of mass of blocks AA and BB after 2s2s from the application of force
Let the tension in the string be TT , then;
F2T=6aF - 2T = 6a ……….. (1)\left( 1 \right)
T=4×2a\Rightarrow T = 4 \times 2a
T=8a\Rightarrow T = 8a ……….. (2)\left( 2 \right)
Substituting equation (2)\left( 2 \right) in equation (1)\left( 1 \right) we get
F16a=6aF - 16a = 6a
Therefore, a=F22a = \dfrac{F}{{22}}
Substituting the value of FF in above equation we get acceleration as
a=1ms2a = 1\,m{s^{ - 2}}
Then the acceleration of the centre of mass will be
acm=(6×acceleration of A)+(4×acceleration of B)6+4{a_{cm}} = \dfrac{{\left( {6 \times acceleration{\text{ }}of{\text{ }}A} \right) + \left( {4 \times acceleration{\text{ }}of{\text{ B}}} \right)}}{{6 + 4}}
acm=(6×1)+(4×2)10{a_{cm}} = \dfrac{{\left( {6 \times 1} \right) + \left( {4 \times 2} \right)}}{{10}}
acm=1.4ms2\therefore {a_{cm}} = 1.4\,m{s^{ - 2}}

Hence, option A is correct.

Note: By using acceleration of the centre of mass we can find speed of centre of mass that is speed of centre of mass = acceleration ×\times time =1.4×2=2.8ms1 = 1.4 \times 2 = 2.8m{s^{ - 1}} because the acceleration is defined as rate of change of velocity. Is is a vector quantity and its S.IS.I unit is ms2m{s^{ - 2}}.