How Far Up an Incline Does a Released Spring-Compressed Object Travel?

[Jtappan]

Homework Statement

A spring with k = 40.0 N/m is at the base of a frictionless 30.0° inclined plane. A 0.50 kg object is pressed against the spring, compressing it 0.4 m from its equilibrium position. The object is then released. If the object is not attached to the spring, how far up the incline does it travel before coming to rest and then sliding back down? (See the figure below.)
m

Homework Equations

i have no idea :-(

The Attempt at a Solution

I can do a regular spring constant problem without the incline. How do i do it with the incline when the height of the incline or the distance of the incline are not given?

 

[Anadyne]

Since the floor is frictionless, it doesn't matter how far away the spring is. You don't need the distance of the incline or the height. (You can assume it's big enough so the block doesn't fall off or anything like that).

All the spring does is give the block some initial velocity at the base of the incline. Remember the "good old days" when you just threw up a ball with some initial velocity and you had to figure out the maximum height? It's the same thing as soon as the block isn't in contact with the spring anymore. Once you find the height you can find the distance up the incline with trig.

 

[ogb p]

As a scientist, it is important to approach problems with a systematic and logical method. In this case, we are dealing with an inclined plane and a spring, so we can use principles of mechanics and energy conservation to solve this problem.

First, let's draw a free body diagram of the object on the inclined plane. We can see that there are two forces acting on the object: the weight of the object (mg) and the normal force from the inclined plane (N). Since the plane is frictionless, there is no friction force.

Next, we can use Newton's second law to write an equation of motion for the object:

ΣF = ma

Where ΣF is the sum of all forces acting on the object, m is the mass of the object, and a is the acceleration of the object. In this case, the only force acting on the object along the incline is the component of the weight force (mg sinθ) and the normal force (N) is perpendicular to the incline, so it does not contribute to the motion along the incline. Therefore, our equation becomes:

mg sinθ = ma

We can also use Hooke's law to relate the force from the spring to the displacement of the object:

F = -kx

Where F is the force from the spring, k is the spring constant, and x is the displacement from the equilibrium position. Since the object is compressed by 0.4 m, the force from the spring can be calculated as:

F = -kx = -(40.0 N/m)(0.4 m) = -16 N

Now, we can substitute this force into our equation of motion:

mg sinθ = ma = -16 N

We also know that the acceleration (a) is equal to the second derivative of the displacement (x) with respect to time (t):

a = d^2x/dt^2

Therefore, we can rewrite our equation as:

mg sinθ = m(d^2x/dt^2) = -16 N

We can solve for the second derivative of the displacement by rearranging the equation:

d^2x/dt^2 = -(16 N)/m = -(16 N)/(0.50 kg) = -32 m/s^2

Now, we can integrate this equation twice to find the displacement (x) as a function of time (t):

x(t) = (-32 m/s