Sandra
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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