Apply the law of conservation of energy to an object

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SUMMARY

The discussion focuses on applying the law of conservation of energy to an object launched upward in Earth's gravitational field. It emphasizes that in the absence of nonconservative forces, the total mechanical energy remains constant, represented by the equation K_i + U_1 + W_other = K_2 + U_2. Participants clarify the correct expressions for kinetic energy (K) and gravitational potential energy (U) to solve for the object's speed at various heights. The discussion highlights the importance of accurately defining initial and final states in energy transformations.

PREREQUISITES
  • Understanding of kinetic energy (K = 1/2 mv^2)
  • Knowledge of gravitational potential energy (U = mgh)
  • Familiarity with the law of conservation of energy
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation and application of the conservation of energy principle in physics
  • Learn how to calculate gravitational potential energy at different heights
  • Explore the effects of nonconservative forces on mechanical energy
  • Investigate energy transformations in projectile motion
USEFUL FOR

Students studying physics, educators teaching energy concepts, and anyone interested in understanding the mechanics of objects in gravitational fields.

annabelx4
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Homework Statement



Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.

In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.

In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object's kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation

K_i + U_1 + W_other = K_2 + U_2 ,

where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.


What is the speed of the object at the height of ?
Express your answer in terms of and . Use three significant figures in the numeric coefficient.

Homework Equations



K_i + U_1 + W_other = K_2 + U_2

The Attempt at a Solution



(1/2) mv^2 + 0 + 0 = (1/2) mv^2 + mg (v^2 / 4g)

so when I solve for v it = 0

mv^2 = 2(0)

v = 0

What did I do wrong?
 
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So you're assuming it launches from the ground, so Ki=1/2*mVi^2

then at some other point it will have slowed down of course, so Kf=1/2*m*Vf^2, and the potential energy will be U=mgh

You tried to solve for h as a function of its velocity(which at that point will be the same Vf) as in the kinetic energy equation but I don't believe you did it right
 

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