Question

A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed

Answer 1

Answer:

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

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

$a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2$

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

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

Answer 2

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

To solve this question, we'll begin by calculating the acceleration of the block.

How to determine the acceleration

• Spring constant (K) = 3500 N/m
• Mass (m) = 4 Kg
• Compression (e) = 0.2 m
• Acceleration (a) =?

F = Ke

Also,

F = ma

Thus,

ma = Ke

Divide both side by m

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

How to determine the speed

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 175 m/s²
• Distance (s) = 1.3 m

• Final velocity (v) =?

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

Take the square root of both side

v = √455

v = 21.33 m/s

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