Max height at the top of the ramp of object launched by spring

In summary, a student is seeking confirmation on the solution to a physics problem involving a frictionless horizontal platform and a ramp with kinetic friction. The object is launched by a spring and the student uses the equation for conservation of energy to solve for the height reached by the object. However, there may be a mistake in the calculation of the work done by friction.
  • #1
andrewwg94
2
0

Homework Statement


Ok so I had this problem on my quiz today and I am really anxious to see if i got it right.
It went something like this:

You have a frictionless horizontal platform and a ramp with kinetic friction of 0.3. The ramp is 20 degree of the ground. A 2 kg block/object is launched by a spring with k=1000 N/m at the beginning of the horizontal ramp. The spring is stretched back 20 cm. So the idea is that it is launched on the horizontal platform before eventually going up the ramp. use 9.8 for g.


Homework Equations



I used KEi + PEi - Wfriction = KEf+PEf

The Attempt at a Solution


I basically found the KE due to the spring to be 20N. Then using that i calculated work lost due to friction. P. Energy at top of ramp = mgh. so i solved for h.

I got h=.7385

Everything made sense when i did it. I don't know if that's right though.
 
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  • #2
Welcome to PF,

andrewwg94 said:
I basically found the KE due to the spring to be 20N. Then using that i calculated work lost due to friction. P. Energy at top of ramp = mgh. so i solved for h.

Did you mean 20 J (joules)? Energy is measured in joules, not Newtons. If so, this is correct. Your equation, which expresses the basic physical concept behind this problem (namely the conservation of energy) is also correct. Energy is conserved, hence:

KEi + PEi = KEf + PEf + Wfric

Initially, the object is not moving, hence KEi = 0. All of the energy is in the form of elastic potential energy (energy stored in the spring). So PEi = (1/2)kx2 = 20 J.

At the very end, when the object has traveled as high as it is going to go, it is also not moving. Hence, KEf = 0. All of these 20 joules have been converted into some combination of gravitational potential energy and heat (the latter being equal to the work done by friction).

20 J = mgh + Wfric

Wfric = Ffric * d

where Ffric is the frictional force, and d is the distance traveled along the ramp by the object. The distance d is related to h by simple trigonometry, which means that you can express everything in terms of one variable: h.

For the frictional force, Ffric = μFN, where FN is the normal force acting on the object. The normal force always acts perpendicular (i.e. normal) to the two surfaces that are in contact. So, in this case, the normal force is perpendicular to the ramp surface, which means that you need to use the geometry (specifically the inclination angle of the ramp) to figure out what it is. You can figure out what it is by noting that all forces perpendicular to the ramp have to balance each other out, so the normal force is just equal (in magnitude) to the component of the weight that acts perpendicular to the ramp.

So, it looks like you may have been doing well in the beginning, but then did not compute Wfric properly, at least based on what you posted above.
 

1. What factors affect the maximum height of an object launched by a spring?

The maximum height of an object launched by a spring is affected by several factors, such as the amount of force applied by the spring, the mass of the object, the angle at which the object is launched, and the surface of the ramp.

2. How does the force of the spring affect the maximum height of the object?

The force of the spring directly affects the maximum height of the object. The greater the force applied by the spring, the greater the potential energy stored in the spring, resulting in a higher launch velocity and ultimately, a higher maximum height.

3. What is the relationship between the mass of the object and its maximum height?

The mass of the object also plays a role in determining the maximum height. The heavier the object, the more energy is required to launch it to a certain height. This means that a lighter object will generally reach a higher maximum height than a heavier object when launched by the same spring.

4. Does the angle of the ramp affect the maximum height of the object?

Yes, the angle of the ramp does have an impact on the maximum height of the object. The steeper the ramp, the greater the launch velocity of the object, resulting in a higher maximum height. However, too steep of an angle can also cause the object to overshoot and not reach its maximum height.

5. How does the surface of the ramp affect the maximum height of the object?

The surface of the ramp can affect the maximum height of the object in two ways. A rough surface will cause more friction, slowing down the object and resulting in a lower maximum height. On the other hand, a smooth surface will reduce friction and allow the object to reach a higher maximum height. Additionally, a bumpy surface may cause the object to bounce, potentially impacting its maximum height.

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