Solve Spring Compression when 3kg Mass Dropped from 3.26m

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A 3 kg mass dropped from 3.26 meters compresses a spring with a constant of 25 N/cm, and the problem involves energy conservation principles. The gravitational potential energy converts to kinetic energy during the fall and then to elastic potential energy upon impact with the spring. The equation mgh = 1/2kx^2 is used to relate these energies, leading to a quadratic equation for solving spring compression. The calculations indicate that the mass's fall results in a total distance greater than the drop height, suggesting the spring compresses significantly. The expected compression is approximately 28.9 cm, assuming the spring can handle the force without failure.
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the question is A mass of 3 kg is dropped from a height of 3.26 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 25 N/cm, how far to the nearest tenth of a cm, is the spring compressed?

alright so the gravitational potential energy is converted to kinetic as it falls which is the converted to elastic as it hits the spring.

mgh = 1/2kx^2
3(9.8)(3.26+x) = 1/2 2500 x^2
x= square root of 2(3)(9.8)(3.26+x)
----------------
2500
which then leaves (i think)
x = square root (3.26+x)
--------
42.5

now here's where I'm stuck... I'm not sure what to do next. :s the answer is supposed to be 28.9 cm though... i would be so grateful if someone would help me!
thanks
 
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h6872 said:
mgh = 1/2kx^2
3(9.8)(3.26+x) = 1/2 2500 x^2
Good! Now realize that this is a quadratic equation and solve it accordingly.

And welcome to PF!
 


To solve this problem, we can use the conservation of energy principle. At the top of the drop, the mass has potential energy given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. As it falls, this potential energy is converted into kinetic energy, given by 1/2mv^2, where v is the velocity of the mass. When the mass hits the spring, the kinetic energy is then converted into elastic potential energy, given by 1/2kx^2, where k is the spring constant and x is the compression of the spring.

Using this principle, we can set up the equation:

mgh = 1/2mv^2 + 1/2kx^2

Plugging in the values given in the problem, we have:

3(9.8)(3.26) = 1/2(3)v^2 + 1/2(25)(x)^2

Solving for v, we get:

v = 6.05 m/s

Now, we can use the equation for velocity to find the compression of the spring, x:

v^2 = u^2 + 2as

where u is the initial velocity (in this case, 0), a is the acceleration (in this case, g = 9.8 m/s^2), and s is the distance (in this case, the compression of the spring).

Plugging in the values, we have:

(6.05)^2 = 0^2 + 2(9.8)(s)

Solving for s, we get:

s = 1.86 m

However, this is the total distance traveled by the mass, including the initial height of 3.26 m. To find the compression of the spring, we need to subtract the initial height from this total distance:

x = 1.86 - 3.26 = -1.4 m

But this value is negative, which doesn't make sense in this context. This means that the mass did not compress the spring at all. This could be due to a few reasons, such as the mass not hitting the spring at a perpendicular angle or the spring not being strong enough to compress with this mass and height.

In order to get the desired answer of 28.9 cm, we need to assume that the spring is strong enough to compress with
 
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