Solving for k, we get k = 2mgd(sinθ)/x2.

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Homework Help Overview

The problem involves a car of mass m rolling down a frictionless incline at an angle θ and coming to a stop by compressing a horizontal spring over a distance x. The objective is to determine the spring constant k required to achieve this stop.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the energy conservation approach, questioning the initial conditions such as the initial velocity of the car. There is an exploration of the relationship between potential energy at the top of the incline and the energy stored in the spring.

Discussion Status

Some participants provide guidance on interpreting the problem statement, particularly regarding the initial conditions of the car's motion. There is an ongoing examination of the assumptions made about initial velocity and energy conservation principles.

Contextual Notes

Participants note the importance of carefully reading the problem statement to determine initial conditions, as well as the potential for confusion with similar scenarios involving different initial velocities.

flyingpig
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Homework Statement



A car with mass m rolls d down a frictionless \theta^0 degree incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the car in a distance of x?

The Attempt at a Solution



[PLAIN]http://img4.imageshack.us/img4/8837/unledxrq.jpg

I am honestly very embarrassed that I can't do these basica problems properly and I am a rising sophomore.

I tried

\sum W = \Delta K

mg(d\sin\theta) - \int_{x = 0}^{x = x} kx dx = -\frac{1}{2}mv_0^2

I thought that at the top before it collides that

-\frac{1}{2}mv_0^2 = -mg(dsin\theta)mg(d\sin\theta) - \int_{x = 0}^{x = x} kx dx = -mg(d\sin\theta)
\int_{x = 0}^{x = x} kx dx = 2mg(d\sin\theta)k = \frac{4mgd\sin\theta}{x^2}

Apparently this is wrong and the answer is just mgh = 0.5kx^2.

I decided to not use LaTeX for the correct answer because I am upset.

If assume initial velocity is 0 I get it right
k = \frac{2mgd\sin\theta}{x^2}

Why? How do I know it started with 0 velocity?
 
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hi flyingpig! :smile:
flyingpig said:
mg(d\sin\theta) - \int_{x = 0}^{x = x} kx dx = -\frac{1}{2}mv_0^2

yes :smile:

(except you really ought to use a different variable inside the ∫, maybe x' or y)
I thought that at the top before it collides that

-\frac{1}{2}mv_0^2 = -mg(dsin\theta)

i don't understand where this comes from :confused:
 
It came from at the top of the ramp.
 
flyingpig said:
Why? How do I know it started with 0 velocity?

This is the simple part.

Read the question and look for the following:

An object falls, an object is dropped, a car rolls down a hill, a block slides down a slope, ...

In all these cases the object is starting with zero velocity.

When the question starts

An object is thrown, A projectile is lauched, or similar, it is NOT starting with a velocity of Zero.

Actually "an object is dropped/falls" can be tricky, since it can sometimes be dropped/fall from a moving point - but then you will realize it is moving so it shouldn't be a problem.
Examples could be:
You dropped a book while in a lift traveling up at 2 m/s
A bomb falls from an airplane traveling at ...
A bag of nuts falls out of a roller coaster when it speeds over a crest ...

Good luck

Peter
 
At the top of the ramp, the car has PE = mgd(sinθ) and KE = 0.

At the point where the spring has stopped the car, the car has PE = (1/2)k(x2) and KE = 0.

Therefore,
mgd(sinθ) = (1/2)k(x2) .​
 

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