Determining Speed of a Block Launched by a Spring

How to calculate the speed of a block launched by a spring?

Given a spring with a stiffness of 3500 N/m that launches a 4 kg block straight up in the classroom with an initial compression of 0.2 m and the block initially at rest, what is the speed of the block when it is 1.3 m above its starting position?

Answer:

The speed of the block at the given position is 21.33 m/s.

To solve this problem, we need to first calculate the acceleration of the block using the given spring constant and mass of the block. Then, we can determine the speed of the block at a height of 1.3 m above the starting position.

Calculating Acceleration:

Given:

Spring constant (K) = 3500 N/m

Mass (m) = 4 kg

Compression (e) = 0.2 m

Acceleration (a) =?

Using the formula F = Ke and F = ma:

ma = Ke

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

Calculating Speed:

Given:

Initial velocity (u) = 0 m/s

Acceleration (a) = 175 m/s²

Distance (s) = 1.3 m

Final velocity (v) =?

Using the equation:

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

v = √455

v ≈ 21.33 m/s

Therefore, the speed of the block at a height of 1.3 m above the starting position is 21.33 m/s.

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