Question

Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]​

Answer 1

Answer:

Explanation:

F = ma

a = F/m

a = mBg / (mB + mA)

a = 3.5(9.8)/(3.5 + 3.4)

a = 4.971014...

a = 5.0 m/s²

If you want to use individual Free Body Diagrams

mass A will have downward weight and upward normal forces equal at mAg

and a horizontal force of string tension T

F = ma

T = mAa

mass B will have a downward force of mBg and an upward force of T

mBg - T = mBa

substitute for T

mBg - mAa = mBa

mBg = a(mB + mA)

a = mBg / (mB + mA)   which is identical to the above answer.