A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.202 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Question

Jakob Holland

Physics

21 July, 09:36

A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.202 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

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Answers (1)

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Dino · Answer 1

Answer: The block of 0.4 Kg travel the same distance that the block of

0.2 Kg

Explanation: Considering the second newton law, we have the following

F = m*a

F = P*sin (θ) where θ is the angle for the incline

so mg sin (θ) = m*a

a=g sin (θ)

both block have the same acceleration in the inclined plane so travel the same distance independent of its mass.