Block in an incline connected to a spring

[Warmacblu]

Homework Statement

Supposed the incline is frictionless for the system. The angle of inclination is 28 degrees, the spring constant is 692 N / m and the mass of the block is 4.95 kg. The block is released from rest with the spring initially unstretched. The acceleration of gravity is 9.8 m/s².

How far x does it move down the incline before coming to rest?

Choosing down the incline as the positive direction, what is its acceleration at its lowest point?

Homework Equations

For part A - mgh = 1/2kx²

h = xsin28

The Attempt at a Solution

I plugged h into the equation and got:

mg(xsin28) = 1/2kx²

I cannot seem to solve for x.

I do not know if this equation is correct.

Thanks for any help.

 

[TVP45]

You have to be careful, when doing inclines, to keep your directions clear and separate. I recommend using x and y for the untilted world and x' and y' for the incline.

 

[Warmacblu]

I don't entirely understand what you are saying but when I drew out the diagram I set the positive x direction pointing down with the block moving down the incline and I don't even think a y direction has to be taken into account.

 

[TVP45]

Perhaps I misunderstood your setup. You have x being positive down the incline. Where is h? You might also say something about the level of the course this is for since the answer is more mathematically involved than the question seems to suggest.

 

[Warmacblu]

Here is a link to the exact question I am doing but I wanted to try and solve it for myself before looking at this thread:

https://www.physicsforums.com/showthread.php?t=212572

I did glance over it and I got the same equation for the first part but he seemed to be able to solve for x and I could not. I think it's an algebra problem.

 

[Doc Al]

The Attempt at a Solution

I plugged h into the equation and got:

mg(xsin28) = 1/2kx²

I cannot seem to solve for x.

I do not know if this equation is correct.

The equation is fine. Try dividing both sides by x.

 

[Warmacblu]

The equation is fine. Try dividing both sides by x.

mg(xsin28) = 1/2kx²
mg(sin28) = 1/2kx
mg(sin28) / 1/2k = x

Does that look okay?

 

[Doc Al]

mg(xsin28) = 1/2kx²
mg(sin28) = 1/2kx
mg(sin28) / 1/2k = x

Does that look okay?

Looks good.

 

[Warmacblu]

I am going to try out this answer in a few minutes, I will report back.

 

[Warmacblu]

Okay, my x value (distance moving down the incline) turned out to be .066 m. I managed to solve b by using a Fnet equation and an acceleration equation given by my professor. My Fnet equation looked like this:

Fnet = mgsin22 + kx

After solving for F, I plugged that into F / m (mass) = a

However, my answer should be negative because it is moving up the slope at its lowest point.

Thanks for the help on part a.

 

[Doc Al]

Okay, my x value (distance moving down the incline) turned out to be .066 m.

Good.

I managed to solve b by using a Fnet equation and an acceleration equation given by my professor.

That "acceleration equation" is just Newton's 2nd law.

My Fnet equation looked like this:

Fnet = mgsin22 + kx

Careful with the direction (and thus the sign) of those forces. Which way does gravity act? Which way is the spring pulling?

(Also, is 22 degrees a typo? Before it was 28.)