Speed of a ball launched from a spring

[cilantrone]

Homework Statement

The force required to compress an imperfect horizontal spring an amount x is given by F = 150x + 12x^3, where x is in meters and F in Newtons.
If the spring is compressed 2.5m, what speed will it give to a 3.5kg ball held against it and then released?

Homework Equations

Not sure all these are relevant, but...
F = ma
a = dv/dt
K = mv^2 /2

The Attempt at a Solution

I made many attempts and the computer told me that none were right.

Thanks!

 

[Hootenanny]

Homework Statement

The force required to compress an imperfect horizontal spring an amount x is given by F = 150x + 12x^3, where x is in meters and F in Newtons.
If the spring is compressed 2.5m, what speed will it give to a 3.5kg ball held against it and then released?

Homework Equations

Not sure all these are relevant, but...
F = ma
a = dv/dt
K = mv^2 /2

The Attempt at a Solution

I made many attempts and the computer told me that none were right.

Thanks!

Welcome to PF cilantrone,

Just looking at the relevant equations that you have listed, I notice that there is a vital one missing, the definition of work.

 

[Moetasim]

As a scientist, it is important to approach problems systematically and use the appropriate equations and units to solve them. In this case, we can use the equation F = ma to calculate the acceleration of the ball as it is released from the compressed spring. We can then use the equation a = dv/dt to calculate the velocity of the ball at any given time.

To start, let's convert the given force equation to SI units by converting meters to centimeters and Newtons to grams. This gives us F = 150x + 12x^3 = 150(100x) + 12(100x)^3 = 15000x + 1200000x^3, where x is now in centimeters and F is in grams. We can then use the equation F = ma to find the acceleration of the ball, which is a = F/m = (15000x + 1200000x^3)/3.5 = 4286x + 342857x^3, where x is still in centimeters and a is in cm/s^2.

Next, we can use the equation a = dv/dt to find the velocity of the ball at any given time. Since the ball is released from rest, we can integrate this equation to find the velocity at any time t. This gives us v = ∫a dt = ∫(4286x + 342857x^3) dt = 2143x^2 + 85714x^4 + C, where C is the constant of integration.

Now, let's plug in the given value of x = 2.5m (250cm) to find the velocity of the ball at the moment it is released. This gives us v = 2143(250)^2 + 85714(250)^4 + C = 133906250 + C. Since the ball is initially at rest, the constant of integration can be set to 0. Therefore, the velocity of the ball at the moment it is released from the compressed spring is v = 133906250 cm/s.

To convert this to m/s, we can divide by 100 to get v = 1339062.5 m/s. Therefore, the speed of the ball launched from the spring is approximately 1.34 million meters per second. This is an incredibly high speed and would likely cause the ball to travel a great distance before coming to a