Solving a Problem on Spring Energy Inclined Plane

[wetcarpet]

1)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 0.2 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? --I attached the picture, although it's not really necessary due to the description above.--

I started by finding the Potential Energy of the spring:
U = 1/2kx^2
U = 1/2(40)(.2)^2 = .8 J
But, I have no idea where to proceed from that point forward. Can anyone help me out? I know that the answer is .33m, I simply can not understand how I am to arrive at that answer.

 

[quasar987]

It's actually -0.8J. This amount will be transferred to kinetic energy after the block is released. It will then go up the ramp, where it will have to fight against the gravitational force, hence its kinetic energy will be transferred to gravitationnal potential energy. Knowing that U = mgh, you get the height it will attain.

 

[thepatientmental]

To solve this problem, we can use conservation of energy. Initially, the object has only potential energy due to the compressed spring. As it travels up the incline, this potential energy will be converted into kinetic energy. At the highest point, all of the potential energy will have been converted into kinetic energy and the object will come to a momentary stop before sliding back down the incline. We can equate the initial potential energy to the final kinetic energy to solve for the distance traveled up the incline.

Initial potential energy: U = 1/2kx^2 = 1/2(40)(0.2)^2 = 0.8 J

Final kinetic energy: K = 1/2mv^2

Since the object comes to a stop at the highest point, the final velocity will be 0. Therefore, we can set the initial potential energy equal to the final kinetic energy:

0.8 J = 1/2(0.5)v^2

Solving for v, we get v = 2 m/s.

Now, we can use the equation for displacement with constant acceleration to calculate the distance traveled up the incline:

d = v^2/2gsinθ

where g is the acceleration due to gravity and θ is the angle of the incline.

Plugging in the values, we get:

d = (2)^2/(2*9.8*sin30) = 0.33 m

Therefore, the object will travel 0.33 m up the incline before coming to rest and sliding back down.