Inclined ramp, object moving down towards spring

In summary, the force on a crate at the bottom of a ramp that slopes downward at an angle will be equal to (k*1490) + (U*540) - W, where k is the force constant of the spring and U is the elasticity of the spring.
  • #1
UppityHorse
3
0

Homework Statement



You are designing a delivery ramp for crates containing exercise equipment. The crates weighing 1490 N will move at a speed of 2.10 m/s at the top of a ramp that slopes downward at an angle 25.0°. The ramp exerts a kinetic friction force of 540 N on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 7.70 m along the ramp. Once stopped, a crate must not rebound back up the ramp.

Calculate the force constant of the spring that will be needed in order to meet the design criteria.

Homework Equations



I have been using:

[tex]K_{1} + U_{grav1} + U_{elastic1} - W_{friction} = K_{2} + U_{grav2} + U_{elastic2}[/tex]

The Attempt at a Solution



[tex]K_{1}= (1/{2})(1490/{9.8})(2.1^{2})=335.25 J[/tex]
[tex]U_{grav1} = U_{elastic1} = 0[/tex] (because at the top, I have y=0)
[tex]W_{friction} = (540)(7.7) = 4158 J[/tex] (force * distance)

[tex]K_{2} = 0[/tex]
[tex]U_{grav2} = (1490)(-7.7)(cos 25) = -10398.1 J[/tex]
[tex]U_{elastic2} = (1/2)(k)(7.7^{2})[/tex]

So, my first thought is, I have no idea how long this spring is, so I have no clue how much it is being compressed. Would I be correct to say that x=(7.7)^2 in the U_elastic equation?

Using this and solving for the k in the 2nd elastic equation, I am not getting a correct answer. I am getting 221.803.

This is what I input into my calculator, so you can see how I arranged the equation to come up with K:
(335.25-4158+10398.1)*2/(7.7^2)

Please point out my error, thanks a lot.
 
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  • #2
I see one error here:

[tex]U_{grav2}=mgh=(1490)(7.7\sin(25))[/tex], not cos(25).

See if this helps. With regard to the length of the spring, it seems weird that they don't give you any information about it. Are you sure you don't know anything about it? Try to use your current assumption and see what happens now with the new trig function...
 
  • #3
Thanks. What I posted was the entire text of the problem as it was given to me. There isn't any additional information.

So we have (1490)(-7.7)(sin 25) = -4848.7.

If I substitute your correction into the formula,
(335.25-4158+4848.7)*2/(7.7^2), I get 34.6079, which is something I actually tried earlier and that wasn't correct either.

(I assume you meant (1490)(-7.7)sin(25), otherwise when I move it into the left hand side, I get a negative number result for k, -292.51).
 
  • #4
Ok. So, the information that neither of us has used yet is that we want the crate to stay put after it is stopped. This means that the net force on it at this point must be zero. Can you set up an equation describing this? If you can you'll be left with two equations and two unknowns. Can you take it from here?
 

1. How does an inclined ramp affect an object rolling down towards a spring?

As an object rolls down an inclined ramp towards a spring, the ramp's slope and length will determine the object's speed and momentum as it reaches the spring. The steeper the slope and the longer the ramp, the more potential energy the object will have, resulting in a higher velocity upon reaching the spring.

2. What is the relationship between the angle of an inclined ramp and the distance an object travels towards a spring?

The angle of an inclined ramp is directly proportional to the distance an object will travel towards a spring. This means that as the angle of the ramp increases, the distance the object will travel towards the spring will also increase.

3. How does the mass of an object affect its movement down an inclined ramp towards a spring?

The mass of an object has a direct impact on its movement down an inclined ramp towards a spring. Objects with a greater mass will have more inertia, making them move slower down the ramp compared to objects with a lower mass.

4. Can the speed of an object moving down an inclined ramp towards a spring be controlled?

Yes, the speed of an object moving down an inclined ramp towards a spring can be controlled by adjusting the angle and length of the ramp. A steeper and longer ramp will result in a faster-moving object, while a shallower and shorter ramp will slow down the object's speed.

5. How does friction affect an object's movement down an inclined ramp towards a spring?

Friction plays a significant role in an object's movement down an inclined ramp towards a spring. Friction between the ramp and the object will slow down its speed and reduce its potential energy as it reaches the spring. To minimize the effects of friction, a smooth and low-friction surface should be used for the ramp.

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