Question

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

Answer 1

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