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A 40kg slab rests on a frictionless floor. A 10kg block rests on top of the slab. The coefficient kinetic friction between the block and the slab is 0.40. A horizontal force of 100N is applied on the 10kg block. Find the resulting acceleration of the slab. $(g = 10 {ms}^{- 1})$

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1 {ms}^{- 2}

2 {ms}^{- 2}

3 {ms}^{- 2}

4 {ms}^{- 2}

answer is C.

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Detailed Solution

If two blocks move together

$a = \frac{F}{m} = \frac{100}{50} = 2 {m/s}^{2}$

For upper block

$100 - f = 10 a$

$100 - 0 .4 \times 100 = 10a$

$100 - 40 = 10 a$

$a = 6 {m/s}^{2}$

For lower block (slab)

$f = ma$

$40 = 40 a$

$a = 1 {m/s}^{2}$

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