Really on simple energy and spring on incline

In summary, the block placed against a spring on a frictionless 30.0° incline with a spring constant of 19.6 N/cm is compressed 16.0 cm and then released. The elastic potential energy of the compressed spring can be calculated using the formula PE=1/2KX^2, where K is the spring constant and X is the compression distance. To convert the units to joules, the spring constant must be converted to N/m. The correct conversion is 19.6 N/cm * 100 cm/m = 1960 N/m. Therefore, the correct calculation for the elastic potential energy is PE=.5(.196)(.16^2) = .0025088J.
  • #1
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A 2.00 kg block is placed against a spring on a frictionless 30.0° incline (Fig. 8-33). (The block is not attached to the spring.) The spring, whose spring constant is 19.6 N/cm, is compressed 16.0 cm and then released.

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(a) What is the elastic potential energy of the compressed spring?
So I think of it like this, it wants the answer in joules, so I convert 19.6N/cm to .196N/m, and 16cm to .16m anyway. PE=1/2 K X^2
so
PE=.5(.196)(.16^2) = .0025088J but it's wrong. I know this is very simple, But I don't understand why I am wrong here
 

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  • #2
so I convert 19.6N/cm to .196N/m

This is wrong. Try again, make sure your units work out.
 
  • #3
I tried leaving them the same, but that didn't work either. The only other thing I can think of is leaving one the same and changing the other, which is mixing units, and that's wrong. What is wrong about it, can you be a little more specific?
 
  • #4
You just didn't convert it properly. Write it out so you can see how the units need to be:

19.6 N/cm * 100 cm/m = 1960 N/m.
 
  • #5
Ahh thank you, I swear I lose so many points because of stuff like this
 

FAQ: Really on simple energy and spring on incline

What is the relationship between energy and springs on an incline?

The amount of potential energy stored in a spring on an incline is directly proportional to the height of the incline and the spring's stiffness. As the height of the incline increases, so does the potential energy stored in the spring. Similarly, a stiffer spring will have a higher potential energy compared to a less stiff spring on the same incline.

How does the mass of an object affect its potential energy on an incline?

The mass of an object has no effect on its potential energy on an incline. The potential energy is solely dependent on the height of the incline and the stiffness of the spring. This means that two objects of different masses on the same incline with the same spring will have the same potential energy.

Can a spring on an incline have both potential and kinetic energy?

Yes, a spring on an incline can have both potential and kinetic energy. When the spring is compressed, it has potential energy stored in it. As the spring is released, it begins to move and gains kinetic energy. The potential energy decreases as the kinetic energy increases, and vice versa.

How does the angle of an incline affect the potential energy of a spring?

The angle of the incline does not affect the potential energy of a spring. As long as the height of the incline and the stiffness of the spring remain constant, the potential energy will remain the same regardless of the incline's angle.

How can I calculate the potential energy of a spring on an incline?

The potential energy of a spring on an incline can be calculated using the formula PE = 1/2kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. The height of the incline can also be used to calculate potential energy using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the incline.

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