Maximum speed of a box compressing a spring on an inclined p

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SUMMARY

The maximum speed of a box with a mass of 40 kg, released from rest on a frictionless ramp at a 45-degree angle, is calculated using energy conservation principles. After sliding 2.00 m down the ramp and compressing a spring with a force constant of 500 N/m, the box reaches a maximum speed of 1.96 m/s. The calculations involve equating gravitational potential energy to spring potential energy, leading to the conclusion that the displacement of the spring is 0.5547 m. The discussion emphasizes the importance of using consistent energy methods throughout the problem-solving process.

PREREQUISITES
  • Understanding of gravitational potential energy (mgh)
  • Knowledge of spring potential energy (1/2 k x^2)
  • Familiarity with Newton's laws of motion
  • Basic principles of energy conservation
NEXT STEPS
  • Review energy conservation in mechanical systems
  • Learn about the relationship between force and displacement in springs
  • Study the dynamics of motion on inclined planes
  • Explore advanced problems involving energy transformations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of problem-solving in inclined plane scenarios.

harik2000
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Homework Statement



A box of mass m = 40kg is released from rest at thte top of a frictionless ramp. The ramp makes an angle theta = 45 degrees to the horizontal. After sliding fom the rigin a distance d = 2.00 m down the ramp, the box strikes a spring of force constant k = 500 N/m. The box compresses the spring momentarily stops, and then begins to slide back up the ramp. What is the maximum speed of the box?

Homework Equations

The Attempt at a Solution


mgsin(45) = kx
(9.81 * 40)sin45 = 500x
x= 0.5547

1/2(k)(x^2) = 1/2(m)(v^2)
v = 1.96m/s

Can anyone verify this pls?
 
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harik2000 said:
mgsin(45) = kx
(9.81 * 40)sin45 = 500x
x= 0.5547

1/2(k)(x^2) = 1/2(m)(v^2)
v = 1.96m/s

Can anyone verify this pls?

It doesn't look right to me. Any reason why you used force considerations to do the first part, and energy do do the latter part?
 
Mister T said:
It doesn't look right to me. Any reason why you used force considerations to do the first part, and energy do do the latter part?

I was trying to find the displacement provided when the ball compresses the spring the most however realized that I did not use potential energy mgh and I was confused. Can you show me the steps? I am confused.
 
harik2000 said:
I was trying to find the displacement provided when the ball compresses the spring the most however realized that I did not use potential energy mgh and I was confused. Can you show me the steps? I am confused.

What part confuses you?
 
I am confused about the potential energy so it would be equal to mgh = 1/2 k x^2 but how would that help solve for v?
 
harik2000 said:
I am confused about the potential energy so it would be equal to mgh = 1/2 k x^2 but how would that help solve for v?

Solve it for x and tell us what you get.
 
harik2000 said:
I am confused about the potential energy so it would be equal to mgh = 1/2 k x^2 but how would that help solve for v?
It won't.
First, please check you have stated the problem exactly.
Is it the whole problem or are there more parts? If there are more parts (and even if there aren't) consider the possibility that not all the information is relevant to finding the max speed.
Then consider at what point in the whole process you would expect max speed to be reached.
 

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