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Physics · LAWS OF MOTION
हिन्दीA wooden block of mass \( m_1 = 3\text{ kg} \) rests on a soft horizontal floor. An iron cylinder of mass \( m_2 = 17\text{ kg} \) is placed on top of the block. The floor yields steadily, causing the block-cylinder system to accelerate downwards at \( a = 2\text{ m/s}^2 \). Taking \( g = 10\text{ m/s}^2 \), calculate (i) the normal action force exerted by the block on the floor, and (ii) the normal reaction force exerted by the cylinder on the block.
Correct answer · Option 2
\( 160\text{ N} \) and \( 136\text{ N} \)
Explanation
For the total system downwards: \( (m_1 + m_2)g - R_{\text{floor}} = (m_1 + m_2)a \implies 20(10) - R_{\text{floor}} = 20(2) \implies R_{\text{floor}} = 160\text{ N} \). For the cylinder alone downwards: \( m_2 g - N = m_2 a \implies 17(10) - N = 17(2) \implies N = 136\text{ N} \).
