A block dropped on a spring, distance compressed unknown.

AI Thread Summary
A block with a mass of 1.6 kg is dropped from a height of 61 cm onto a spring with a spring constant of 1820 N/m. The discussion revolves around calculating the maximum compression of the spring using energy conservation principles, equating gravitational potential energy to elastic potential energy. Attempts to solve the problem led to complications, including negative values in the quadratic equation derived from the energy equations. Participants emphasized the importance of unit consistency and proper equation formatting to avoid errors. The conversation highlights the challenges faced in applying conservation laws to determine the spring's compression accurately.
ramses07
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A block of mass m = 1.6 kg is dropped from height h = 61 cm (height above the spring, not total height) onto a spring of spring constant k = 1820 N/m (Fig. 8-38). Find the maximum distance the spring is compressed.


Homework Equations



K=-U, since all energy is conserved, work would equal zero.

The Attempt at a Solution



basically kinetic energy is zero, since its starts from zero and ends with zero.

so i this means that the potential energy must equate.

mgh (potential energy from block at H) = (kx^2)/2(potential energy from spring)

so basically the height from the spring is disclosed, but the spring compression isnt.

I tried this first

mg(x+h)=1/2kx^2

1.6kg * 9.8 m/s^2(x+.61cm)=910n/m *x^2


but it gives me an awkward quadratic equation, where the number under the square root would be negative.

I also tried to find to find the kinetic energy right before it touches the spring, and then apply it to the conservation of energy, where 1/2mvf^2=1/2kx^2 but that ddidnt work either, any suggestions?
 
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ramses07 said:
I also tried to find to find the kinetic energy right before it touches the spring, and then apply it to the conservation of energy, where 1/2mvf^2=1/2kx^2 but that ddidnt work either...

Pls elaborate. What didn't work? Be careful with units (i.e. cm vs m).
 
well basically i found the kinetic energy right as it hits the spring, by using the equation
K=U.

which would be 1/2mvf^2-1/2mvi^2=-mgh

the initial velocity would be zero, so that cancels, and leaves me with

1/2mvf^2=-mgh

so i use the kinetic energy produced by the distance from the spring and plug it into the
conservation of elastic energy equation

-1/2mvi^2(kinetic energy produce by the drop)=1/2kx^2(potential elastic energy)

but this equation didnt produce the correct answer either, and i actually don't know the answer since its an online homework assessment.
 
Looks good so far. What is your expression for x and why do you think the answer you got is wrong? What answer did you get by the way?
 
Last edited:
ramses07 said:
1.6kg * 9.8 m/s^2(x+.61cm)=910n/m *x^2

but it gives me an awkward quadratic equation, where the number under the square root would be negative.

Why would it give you a negative number? Did you put the equation to the form of ax^2+bx+c=0 before you start? -4ac is a positive number because a is negative and c is positive.
 
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