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Q.

A block of mass 0 . 5 kg has an initial velocity of 10 m/s down an inclined rough plane of angle 300 as shown in fig. (6). The coefficient of friction between the block and inclined surface being 0 2. The velocity of the block after it travels a distance of 10 m is

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a

2a m/s

b

13 m/s

c

8 m/s

d

17 m/s

answer is B.

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Detailed Solution

The acceleration is given bya=gsin⁡30∘−μcos⁡30∘ =9⋅812−0⋅2×32=9⋅8×(0⋅3268) =3⋅2m/s2  Now v2=u2+2as v2=(10)2+2×3⋅2×10=164 v≈13m/s
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