Test Index

Laws of Motion

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Question : 32 of 40
Marks: +1, -0
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?
Solution:  
Here, mass of block, m=25 kgm = 25\,\text{kg}
Mass of man, M=50 kgM = 50\,\text{kg}
Force applied to lift the block,
F=mg=25×10=250 N.F = mg = 25 \times 10 = 250\,\text{N}.
Weight of man, W=Mg=50×10=500 NW = Mg = 50 \times 10 = 500\,\text{N}
(a) When block is raised by man as shown in figure (a), force is applied by the man in the upward direction.
Action on the floor by the man =W−F=500+250=750 N= W - F = 500 + 250 = 750\,\text{N}
(b) When block is raised by man as shown in figure (b), force is applied by the man in the downward direction.
Action on the floor by the man =W−F=500−250=250 N= W - F = 500 - 250 = 250\,\text{N}
As the floor yields to a normal force of 700 N, the mode (b) has to adopted by the man to lift the block.
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