Energy Conservation: Block & Spring

AI Thread Summary
An object of mass 4 kg slides down a frictionless incline and compresses a spring by 1 m after losing gravitational potential energy. The initial potential energy lost when the block touches the spring is calculated to be 53.7 J, while the total loss of gravitational potential energy after full compression is 67.1 J. The spring constant is derived from the energy conservation principle, where all potential energy is converted into spring energy. The initial attempt to calculate the spring constant incorrectly subtracted the two energy values instead of applying conservation of mechanical energy. The correct approach confirms that the total gravitational potential energy loss is equal to the spring energy at maximum compression.
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Homework Statement


An object of mass M = 4 kg slides from rest a distance d = 4 m down a frictionless inclined plane where it encounters a spring. It compresses the spring a distance D = 1 m before stopping. The inclined plane makes an angle q = 20° with the horizontal.

Homework Equations


When the block just touches the spring, how much gravitational potential energy has it lost?
U=53.7 J
After the mass has fully compressed the spring, what is its total loss of gravitational potential energy?
U=67.1 J
What is the value of the spring constant?
k=??

The Attempt at a Solution


U=Ui-Uf
U=53.7-67.1=-13.4
U=(1/2)kx^2
k=2U/x^2
k=2(-13.4)/(5)^2
k=1.07?
the answer is wrong. am i suppose to use the conservation of mechanical energy? because i am lost.
 
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It is wrong to take the difference between 53.7 and 67.1.
The ramp is frictionless, so all of the PE (67.1 J) goes into the spring energy.
 
thank you, i figured it out.
 
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