Two blocks of mass 10 kg and 4 kg are placed over the surface of a lift accelerating upward with a=2m/sec^2 contact force between the block is equal to
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Answer:
The answer is 20N.
Explanation:
The formula of the contact force is
F-f = m1a
Now let's find the force (F) acting on the system.
Given data:
m1 = 4 Kg
m2 = 10 Kg
a = 2 m/ sec^2
a = F/ m1 + m2
a (m1 + m2) = F
F = a (m1 + m2)
F = 2 (4 +10) = 2 x 14
F =28 N
Put the vlaues:
F-f = m1a
f = F - m1a
f = 28 - 4 x 2
f = 28 - 8 = 20 N
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