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#### See Fig. 5.15. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m/s^2 . What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m/s^2 . Identify the action-reaction pairs in the problem.

145 viewed last edited 3 months ago An0nym0us
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Given that

Mass of wooden block m = 2 kg

Mass of iron cylinder M = 25 kg

Acceleration of the block and cylinder a = 0.1 m/s^2

Acceleration due to gravity g = 10 m/s^2

a)

The earth's gravitational pulling force and the normal force R exerted by the block's floor are both acting on the block on the floor. According to the newton's third law of motion

Action: the block's force on the floor

Reaction: the floor's force on the block

R=mg

R=2*10

R=20N

Thus, force exerted on the floor by the block ﻿ R = 20 N

b) Total mass of the system = mass of the block + mass of the cylinder

m_s = m + M = 2 kg + 25 kg = 27 kg

Forces acting on the system

(i) System is accelerating downwards

(ii) Normal force acting upwards

(ii) Weight of the system acting downwards

\therefore Action force F = m_s g - R

m_s a = m_s g - R

R = 27 *10 - 27 *(0.1)

R = 267.3

Hence, the action of the system on the floor R = 267.3 N