Conservation of mechanical energy question

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SUMMARY

The discussion centers on calculating the maximum distance a block of mass 2.5 kg travels up an inclined plane after being released from a compressed spring with a spring constant of 2900 N/m. The calculations utilize the conservation of mechanical energy principle, specifically the equations E = 1/2kx² and E = mgh. The user correctly finds the height (h = 1.16 m) and subsequently calculates the distance (d = 2.47 m) using the incline angle of 28 degrees. Feedback indicates that while the approach is generally correct, the energy equations used need to be consistent for accurate results.

PREREQUISITES
  • Understanding of mechanical energy conservation principles
  • Familiarity with spring potential energy calculations
  • Knowledge of gravitational potential energy equations
  • Basic trigonometry for calculating distances on inclined planes
NEXT STEPS
  • Review the conservation of mechanical energy in systems involving springs and inclined planes
  • Learn about the effects of friction on energy calculations, specifically with kinetic friction (µk = 0.22)
  • Explore advanced applications of the work-energy theorem in physics problems
  • Practice solving similar problems involving different masses and spring constants
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jgray
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Homework Statement



A block of mass 2.5 kg is placed against a compressed spring (k = 2900 N/m) at the bottom of an inclined plane (angle = 28 degrees). When the spring is released the block is projected up the incline and the spring expands by 14 cm to its normal length.
a.Calculate the maximum distance traveled by the block up the incline without friction.
b.Repeat the calculation with friction, taking µk = 0.22.

Homework Equations



E=1/2kx^2
E=1/2mv^2
E=mgy + 1/2kx^2

The Attempt at a Solution



Stuck on part a since I'm not sure its right.. :
E1=E2
mgh=1/2kx^2
(2.5)(9.8)h=1/2(2900)(.14)^2
24.5h=28.42
h=1.16m
So since I have one side of the triangle and the angle, I should be able to find the distance:
d= h/sin28 = 1.16/0.469471562= 2.47 m

Am I on the right track so far?? Thanks for any advise!
 
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jgray said:

Homework Statement



A block of mass 2.5 kg is placed against a compressed spring (k = 2900 N/m) at the bottom of an inclined plane (angle = 28 degrees). When the spring is released the block is projected up the incline and the spring expands by 14 cm to its normal length.
a.Calculate the maximum distance traveled by the block up the incline without friction.
b.Repeat the calculation with friction, taking µk = 0.22.

Homework Equations



E=1/2kx^2
E=1/2mv^2
E=mgy + 1/2kx^2

The Attempt at a Solution



Stuck on part a since I'm not sure its right.. :
E1=E2
mgh=1/2kx^2
(2.5)(9.8)h=1/2(2900)(.14)^2
24.5h=28.42
h=1.16m
So since I have one side of the triangle and the angle, I should be able to find the distance:
d= h/sin28 = 1.16/0.469471562= 2.47 m

Am I on the right track so far?? Thanks for any advise!
That looks like the right result.

Your energy equations are inconsistent.

PESPRING = (1/2)kx2 .

PEGRAVITY = mgh

E = KE + PE
 

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