How Is Maximum Elongation Calculated in a Spring System?

[MathewsMD]

A 2-kg block is attached to a horizonal ideal spring with a spring constant of 200 N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

Is not the solution simply:

1/2kx² = 1/2mv²

x = (mv²/k)^0.5

x = 0.5 m?

That is incorrect, though.

 

[Doc Al]

Looks correct to me.

 

[rude man]

Me too.

 

[MathewsMD]

The answer was 5m...probably a typo, just wanted to confirm.

Thanks.

 

[Nickie]

The correct solution is as follows:

The maximum elongation of the spring can be found by using the equation for potential energy stored in a spring:

U = 1/2kx^2

Where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

We are given the mass of the block (m = 2 kg), the spring constant (k = 200 N/m), and the initial velocity of the block (v = 5 m/s). We can use the conservation of energy principle to equate the initial kinetic energy of the block to the potential energy stored in the spring at maximum elongation.

1/2mv^2 = 1/2kx^2

Solving for x, we get:

x = √(mv^2/k)

Substituting the given values, we get:

x = √((2 kg)(5 m/s)^2/(200 N/m)) = 0.5 m

Therefore, the maximum elongation of the spring is 0.5 meters. This means that when the block is attached to the spring and given an initial speed of 5 m/s, the spring will stretch by a maximum of 0.5 meters before returning to its equilibrium length.