Spring and object on ramp

In summary, the student made a mistake in calculating the force down the ramp, using mg/sinθ instead of mgsinθ. This resulted in the incorrect answer for the first two questions, but the mistake canceled itself out for the last question, resulting in the correct answer.
  • #1
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I figured out where I was going wrong, it was in the trig for getting the force down the ramp. mgsinθ, not mg/sinθ, although I still got the right answer for the last part using mg/sinθ.

[STRIKE]

Homework Statement



A spring is set to move along the surface of a frictionless ramp, which is tilted at an angle θ = 60 degrees away from horizontal. A rock of mass M = 5 kg is placed on a spring. As a result, the spring compresses, coming to rest a distance x = 3.875 cm along the ramp away from its original position.

What is the spring's force constant?

The rock is pushed down along the ramp an additional distance y = 29 cm. What is the spring's potential energy now?

The rock is now released, so that the spring flings it up along the ramp. The spring and rock lose contact when the rock reaches the spring's rest length. How far along the ramp beyond this release point does the rock slide? (You may assume that the ramp is large enough that the rock never reaches its top end).

The set-up looked like http://spiff.rit.edu/webwork2_course_files/phys216/tmp/gif/set17-prob4-./tilt_rock_spring.gif.

Homework Equations



F=-k/x
SPE=(1/2)kx2
KE=(1/2)mv2
v2=vi2+2ax

The Attempt at a Solution



What is the spring's force constant?
x=0.03875m
Fg=(mg)/sinθ
k=((mg)/sinθ)/x=((5kg*9.8m/s2)/sin60°)/0.03875m≈1460 N/m
The program marked this wrong

What is the spring's potential energy now?
x=0.03875m+0.29m=0.32875m
SPE=(1/2)kx2=(1/2)(1460N/m)(0.32875m)2≈78.9 J
The program also marked this wrong.

How far along the ramp beyond this release point does the rock slide?
v2=2*KE/m=2*78.9 J/5kg≈31.56(m/s)2
x=v2/(2(g/sinθ))=31.56(m/s)2/(2(9.8m/s2/sin60°))≈1.395m
The program marked this correct.

Based on that I'm thinking that either the professor put in the incorrect answers for the first two, or I did something wrong that managed to sort itself out by the last question.[/STRIKE]
 
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  • #2
Zyrn said:
What is the spring's force constant?
x=0.03875m
Fg=(mg)/sinθ
k=((mg)/sinθ)/x=((5kg*9.8m/s2)/sin60°)/0.03875m≈1460 N/m
The program marked this wrong
Check ##F_{g_{x}}##. you calculated it wrong.

$$F_{g_{x}}=-m.g.Sin(θ)$$

Edit: I guess you just figured it out!
 
  • #3
Zyrn said:
I figured out where I was going wrong, it was in the trig for getting the force down the ramp. mgsinθ, not mg/sinθ, although I still got the right answer for the last part using mg/sinθ.
Yep. You made the same mistake twice, and that resulted in the mistake canceling itself out. Every once in a while two wrongs *do* make aright. Don't count in it, however. Two wrongs more often than not compound the problem rather than canceling it out.
 

1. What is a spring and object on ramp experiment?

A spring and object on ramp experiment is a physics experiment that involves studying the motion of a spring attached to an object as it moves down a ramp. The goal of the experiment is to understand the relationship between the applied force, the displacement of the object, and the spring constant.

2. What materials are needed for a spring and object on ramp experiment?

To conduct a spring and object on ramp experiment, you will need a ramp, a spring, an object (such as a toy car or a ball), a meter stick or ruler, a force sensor, and a data collection device.

3. How do you set up a spring and object on ramp experiment?

To set up the experiment, place the ramp at an angle and attach the spring to the top of the ramp. Then, attach the object to the bottom of the spring. Use the force sensor to measure the force applied to the object and the data collection device to record the displacement of the object as it moves down the ramp.

4. What are the key principles behind a spring and object on ramp experiment?

The key principles behind this experiment are Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring, and Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it.

5. How can a spring and object on ramp experiment be applied in real life?

The results of a spring and object on ramp experiment can be applied in various real-life situations, such as understanding the motion of a car on a ramp or calculating the force needed to move an object up a slope. This experiment also helps in understanding the behavior of springs in various mechanical systems, such as shock absorbers and trampolines.

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