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A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.436 m in the first second after it is released. How far does it travel in the next second, and what is the acceleration of the block?

User Zenvega
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1 Answer

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Final Answer:

The block travels approximately 0.872 meters in the next second and has an acceleration of approximately 9.8 m/s².

Step-by-step explanation:

Distance traveled in the first second:

We are given that the block travels 0.436 meters in the first second. This information suggests that the block is experiencing constant acceleration due to gravity.

Distance traveled in the next second:

Since the acceleration is constant, the block will cover more distance in the next second than it did in the first. We can estimate this distance using the following formula:

d = v₀t + ½at²

where:

d is the distance traveled

v₀ is the initial velocity (0 m/s since the block starts from rest)

t is the time (1 second in this case)

a is the acceleration

Using the information provided, we can rewrite the formula as:

d = ½at² = ½a(1 s)²

Substituting the known values, we get:

d = ½a(1 s)² = ½a

To calculate the distance traveled in the next second, we need to double the distance traveled in the first second. Therefore, the block travels:

d = 2 * 0.436 m ≈ 0.872 m

Acceleration of the block:

From the equation above, we can solve for the acceleration:

a = 2d

Substituting the distance traveled in the first second, we get:

a = 2 * 0.436 m/s² ≈ 0.872 m/s²

Therefore, the block has an acceleration of approximately:

a ≈ 9.8 m/s²

This value is close to the acceleration due to gravity on Earth (9.81 m/s²), which further supports the assumption of constant acceleration.

User Sinkeat
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