Two blocks are in contact on a frictionless table one
has a mass m and the other 2m. A force F is applied
on 2m as shown is Figure. Now the same force F is
applied on m. In the two cases respectively the ratio
of force of contact between the two blocks will be-
F2m
m
Answers
Answer:case I
a=F3a
N1=ma
=m×F3a=F3
case II
a=F3a
N2=2m×a
=2m×F3a=23F
N1N2=12
Hence c is the correct answ
Explanation:
Answer:
The ratio of force of contact between the two block will be 2 : 1.
Explanation:
Given:-
Masses, = m & 2m
Force, = F
[Refer to the attached image for the visualization of scenario]
Now, let us consider to be a unit system. Then, we find the acceleration of both the given cases.
[Case 1]
Acceleration, a = F/m
a = F/3m
A normal acts on 2m in this case. So:
Normal, N₁ = ma
N₁ = 2m × F/ 3m
N₁ = 2F / 3 _____(1)
[Case 2]
Acceleration, a = F/m
a = F/3m
A normal acts on m in this case. So:
Normal, N₂ = ma
N₂ = m × F/3m
N₂ = F/3 _______(2)
According to the question, from (1) and (2), we obtain:
N₁ / N₂ = 2F/ 3 / F/ 3
N₁ / N₂ = 2F / 3 × 3/F
N₁ / N₂ = 2 / 1
Hence, the ratio of force of contact between the two blocks will be 2 : 1.
Hope it helps! ;-))
