Three blocks A, B and C of masses 4, 2 and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied onto the 4 kg block, then the contact force between A and B is
medium
Laws of Motion
2015
physics
6 N
8 N
18 N
2 N
Explanation
To solve this problem, we need to find the contact force between blocks A and B.Given:• Mass of block A (mA)=4 kg• Mass of block B (mB)=2
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The blocks are on a frictionless surface, so they will accelerate together.
First, calculate the total mass of the system:
mtotal=mA+mB+mC=4+2+1=7
kg
Next, calculate the acceleration of the system using Newton's second law:
F=mtotal⋅a14=7⋅aa=714=2
m/s
2
Now, consider the forces acting on block B. The contact force
FAB
is the force exerted by block A on block B.
Using Newton's second law for block B:
FAB=mB⋅aFAB=2⋅2=4
N
Therefore, the contact force between blocks A and B is
4
N.
However, since the question asks for the contact force between A and B, we need to consider the force exerted by A on B, which is the same as the force exerted by B on A due to Newton's third law.
