Question

two blocks of mass 10 kg and 4 kg are placed over the surface of a lift accelerating upwards with acceleration 2m/s² contact force between the blocks is equal to​

Answer 1

two blocks of mass $m_1=10kg$ and $m_2=4kg$ are placed over the surface of a lift accelerating upwards with acceleration 2m/s² as shown in figure.

system of block is appeared in non-inertial frame of reference. so, apply pseudo acceleration just opposite direction of motion of lift.

i.e., $a_p$ = 2m/s² [downward]

Let contact force between blocks is F.

from block of mass $m_2$,

at equilibrium,

$m_2g+m_2a_p=F$

or, 4kg × 10m/s² + 4kg × 2m/s² = F

or, F = 40N + 8 N = 48N

hence, contacting force between the blocks is 48N.

Answer 2

Answer:

Explanation:

Mass of block one = m1 = 10 kg (Given)

Mass of block one = m2 = 4 kg (Given)

Acceleration of the lift = 2m/s² (Given)

Let the contact force between the blocks = F.

A pseudo acceleration is always identical to the spectral acceleration. However, in the case of damping other than zero the two are slightly different. Thus, according to the concept of pseudo acceleration -

Thus, from the block of mass = m2 at equilibrium,  

F = 4 × 10 + 4 × 2

F = 40 + 8   

= 48

Therefore, the contacting force between the blocks is 48N.