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A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.692 m in the first second after it is released. How far does it travel in the next second?

Question

A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.692 m in the first second after it is released. How far does it travel in the next second?

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Step 1

Given information:

Distance travelled by the block in the first second (S1) = 0.692 m

Initial velocity (u) = 0 m/s

Let the acceleration of the block be “a”

Step 2

Consider the equation:

1
S utat
2
Substitute the corresponding values:
0.692 a(1)2
We get
a 1.384 m/s2
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1 S utat 2 Substitute the corresponding values: 0.692 a(1)2 We get a 1.384 m/s2

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Step 3

The distance travelled by the body in the next second is given by:

S2 = S0-2 – S1

Where S0-2 is distance travelled by t...

Hence the distance travelled in 2nd second is:
S2 (1.38)(22) - 0.692
S2 2.068 m
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Hence the distance travelled in 2nd second is: S2 (1.38)(22) - 0.692 S2 2.068 m

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Kinematics

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