Question

A lift is moving upward with an acceleration a=0.5ms2. There is a block of mass m-16 kg on the floor of the lift. What is the

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Answer #1

Solution:

Given:

Acceleration (a) = 0.5 m/s2

And, m = 16 kg

Here, Contact force (F) can be given as :

F = m a + m g = m (a + g) = (16 kg){(0.5 m/s2) + (9.8 m/s2)} = 164.8 N

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