2 blocks of masses 2 kg and 1 kg are in contact with each other on A frictionless table when a horizontal force of three N is applied to the block of mass 2 kg the value of force of contact between two blocks is?
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Answered by
5
Answer:
2N
Considering both blocks as a system,
Total mass of the system M = 3 kg
So, net external force for the system = 3 N (as the table is frictionless)
F = Ma
=> 3 = 3a
=> a = 1 ms−2
Now considering the block of 1 kg :
a = 1 ms−2 (acceleration of the 1 kg block is same as the acceleration of the system)
so, F = ma => F = 1 kg x 2 ms−2 = 2 N
Answered by
2
Answer:
The 3.0 N force will cause the total 3 kg to accelerate. The value of that acceleration will be
a = F/m = 3 N / 3kg = 1 m/s²
Since the 1 kg block is accelerating at 1 m/s^2, the force on it must be
F = m×a = 1 kg*1 m/s² = 1 N
a= 3/3= 1m/sec²
T=1×1=1N
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