How Far Up the Incline Does the Object Travel Before Sliding Back Down?

In summary, the conversation is about a problem involving a spring and an inclined plane. The person tried to use a formula to solve the problem but got incorrect answers when submitting to their homework website. They are seeking clarification on what the correct answer should be.
  • #1
vrobins1
22
0

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?


Homework Equations



I tried to use the formula (1/2)kx2=mgh


The Attempt at a Solution



I tried (1/2)(40)(.4)2=.5(9.8)h, solving for h.

The first time I worked it out, I got h=.65306. I tried entering .65, .653, .7 3 separate times into my homework website, and they were all incorrect. I tried working the problem out again, and got h=15.68. I tried submitting 15.7 and 16 as answers, and they were both incorrect as well.
If anyone can offer any insight, it would be greatly appreciated. Thanks!
 
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  • #2
Hi vrobins1,

You were on the right track when you found the first answer for h. However, they want the distance along the incline. What does h represent, and how is that related to the distance traveled along the incline? Once you answer those questions I think you'll get the right answer.
 
  • #3


As a scientist, it is important to first check your calculations and make sure they are correct. In this case, your first attempt at solving for h using the formula (1/2)kx^2=mgh was incorrect. The correct equation to use in this scenario is (1/2)kx^2=mghcosθ, where θ is the angle of the inclined plane. In this case, θ = 30°.

Using this equation, we can solve for h as follows:

(1/2)(40)(0.4)^2 = (0.5)(9.8)hcos30°

h = (0.08 x 2)/(0.5 x 9.8 x cos30°)

h = 0.163 m

Therefore, the object will travel 0.163 m up the incline before coming to rest and sliding back down. It is important to note that this calculation assumes ideal conditions, such as a frictionless inclined plane and no air resistance. In reality, the object may not travel exactly 0.163 m due to factors such as friction and air resistance.
 

FAQ: How Far Up the Incline Does the Object Travel Before Sliding Back Down?

How do springs work?

Springs work by storing potential energy when they are compressed or stretched. This potential energy can then be released as kinetic energy when the spring returns to its original shape.

What factors affect the spring constant?

The spring constant, which determines the stiffness of a spring, is affected by the material, length, and thickness of the spring. It is also affected by the number of coils and the diameter of the wire used to make the spring.

How does the angle of an inclined plane affect the force required to move an object?

The angle of an inclined plane affects the force required to move an object by increasing the effective weight of the object. As the angle increases, the weight component parallel to the inclined plane increases, requiring a greater force to overcome it and move the object.

What is the relationship between the angle of an inclined plane and the distance an object travels?

The angle of an inclined plane is directly proportional to the distance an object travels. This means that as the angle of the inclined plane increases, the distance the object travels also increases.

How does the mass of an object affect its motion on an inclined plane?

The mass of an object does not directly affect its motion on an inclined plane. However, a heavier object will require a greater force to overcome its weight and move up the inclined plane, compared to a lighter object. This is due to the increased weight component parallel to the inclined plane.

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