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Laws of Motion

JEE Main level, single correct, medium

2 of 210

Question 2 of 210

A block $m_1 = 4\,\text{kg}$ on rough horizontal floor ( $\mu_k = 0.2$ ) is connected to hanging mass $m_2 = 6\,\text{kg}$ . System acceleration is ( $g=10\,\text{m/s}^2$ ):

Diagram for this Laws of Motion question

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Answer: (B) $5.2\,\text{m/s}^2$

f_k = \mu m_1 g = 0.2 \times 40 = 8\,\text{N}

. Driving force

= 60\,\text{N}

. Net force

= 52\,\text{N}

. Total mass

= 10\,\text{kg} \implies a = 5.2\,\text{m/s}^2

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Laws of Motion, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A block m1=4 kgm_1 = 4\,\text{kg}m1​=4kg on rough horizontal floor (μk=0.2\mu_k = 0.2μk​=0.2) is connected to hanging mass m2=6 kgm_2 = 6\,\text{kg}m2​=6kg. System acceleration is (g=10 m/s2g=10\,\text{m/s}^2g=10m/s2):

JEE Main question paper PDF

More Physics practice

A block $m_1 = 4\,\text{kg}$ on rough horizontal floor ( $\mu_k = 0.2$ ) is connected to hanging mass $m_2 = 6\,\text{kg}$ . System acceleration is ( $g=10\,\text{m/s}^2$ ):