Question
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 22N is applied on block B when initially the blocks are at rest. Then acceleration of centre of mass of block A and block B, 2s, after the application of force is ( masses of A and B are 4kg and 6kg respectively and surfaces are smooth )
A. 1.4ms−2
B. 1ms−2
C. 2ms−2
D. None of these.
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 B when the blocks are at rest is =22N
F=22N
Mass of block A=4kg
Mass of block A=6kg
We need to find acceleration of centre of mass of blocks A and B after 2s from the application of force
Let the tension in the string be T , then;
F−2T=6a ……….. (1)
⇒T=4×2a
⇒T=8a ……….. (2)
Substituting equation (2) in equation (1) we get
F−16a=6a
Therefore, a=22F
Substituting the value of F in above equation we get acceleration as
a=1ms−2
Then the acceleration of the centre of mass will be
acm=6+4(6×acceleration of A)+(4×acceleration of B)
acm=10(6×1)+(4×2)
∴acm=1.4ms−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 × time =1.4×2=2.8ms−1 because the acceleration is defined as rate of change of velocity. Is is a vector quantity and its S.I unit is ms−2.