Compressing a spring on an angle.

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The discussion centers on calculating the maximum speed of a 10kg box that collides with a spring after being released from a height of 4m on a 30° ramp. The spring has a spring constant (k) of 250 N/m. The key point is that maximum speed occurs when the spring's force equals the component of the box's weight acting down the ramp. The user confirms that the relationship ky = mgsin(30°) can be used to find the compression at which this balance occurs, leading to the solution found in the textbook. Understanding the forces at play during the box's descent and spring compression is crucial for solving the problem.
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Homework Statement


There is a box that is 4m above a spring on a ramp at a 30° angle.
box is 10kg.
we let the box go and it collides with a spring, that has a k of 250N/m
At what compression of the spring does the box have its maximum speed.

The Attempt at a Solution


ok I found the max compression of the spring.
I am not sure how to find the max speed of the box.
Could I just right the speed of the box as a function of spring compression and height and
take its derivative and maximize that?
 

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port31 said:

Homework Statement


There is a box that is 4m above a spring on a ramp at a 30° angle.
box is 10kg.
we let the box go and it collides with a spring, that has a k of 250N/m
At what compression of the spring does the box have its maximum speed.

The Attempt at a Solution


ok I found the max compression of the spring.
I am not sure how to find the max speed of the box.
Could I just right the speed of the box as a function of spring compression and height and
take its derivative and maximize that?

Lets remove the slope for a moment and make this vertical.

When you drop the box, it accelerates towards the Earth, accelerated by a force called its weight, accelerating at 9.8 ms-2.
Once it contacts the spring, and begins to compress that spring, the rate of acceleration reduces. When the spring has been compressed enough that it generates a Force equal to half the weight of the box, the acceleration will have reduced to 4.9 ms-2 - but at least it is still accelerating [gaining speed]
Eventually the spring will be compressed sufficiently to supply an upward force equal to the weight of the box. AT that instant the net force will be zero, and acceleration will be zero.
The box will however continue to move - zero force means there is no reason to stop either - resulting in more compression of the spring, and an even larger upward force from the spring. The net force will be up, the acceleration will be up, the box will be slowing down. SO the maximum speed occurs when the spring has been compressed enough to push back with a force equal to the weight of the box.

On the slope, the probem is similar.
 
ok thanks for the help.
so I just need ky=mgsin(30)
and I get the answer in the back of the book.
 
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