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Second Law of Motion

A 40kg slab...

Question

A $40 k g$ slab rests on frictionless floor. A $10 k g$ block rests on top of the slab. The static coefficient of friction between the block and the lab is $0.60$ while the kinetic coefficient of friction if $0.40$ . The $10 k g$ block is acted upon by a horizontal force of $100 N$ . The resulting acceleration of the slab will be:

A

1.5 m S^{- 2}

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B

2.0 m S^{- 2}

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C

10 m S^{- 2}

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D

0.8 m S^{- 2}

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Solution

The correct option is D $0.8 m S^{- 2}$
${(f_{s})}_{m a x} = μ_{s} \times 10 \times 10 = 60 N 100 N > 60 N = {(f_{s})}_{m a x} H e n c e b l o c k o f 10 k g w i l l m o v e F o r c e o n 40 k g s l a b = f_{k} = μ_{k} \times 10 \times 10 = 40 N a_{s l a b} = \frac{40}{10 + 40} = 0.8 \frac{m}{s e c^{2}}$

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40 kg

slab rests on a frictionless floor. A

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block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient of friction is 0.40. The

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40 k g m

slab rests on a frictionless floor. A

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block rests on the top of the slab. The static coefficient of friction between the block and the slab is

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slab rests on a frictionless floor. A

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block rests on top of the slab. The coefficient of friction between the block and the slab is

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slab rests on a frictionless floor. A

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block rests on top of the slab (as shown in the figure). The coefficient of static friction

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slab rests on frictionless floor. A

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