Part D Using conservation of energy, find the maximum height hmax to which the object will rise. Express your answer in terms of u and g. You may or may not use all of these quantities. IVE ΑΣΦ hmax = Submit Request Answer ?
Kinematics
A machine is a device that accepts energy in some available form and utilizes it to do a type of work. Energy, work, or power has to be transferred from one mechanical part to another to run a machine. While the transfer of energy between two machine parts, those two parts experience a relative motion with each other. Studying such relative motions is termed kinematics.
Kinetic Energy and Work-Energy Theorem
In physics, work is the product of the net force in direction of the displacement and the magnitude of this displacement or it can also be defined as the energy transfer of an object when it is moved for a distance due to the forces acting on it in the direction of displacement and perpendicular to the displacement which is called the normal force. Energy is the capacity of any object doing work. The SI unit of work is joule and energy is Joule. This principle follows the second law of Newton's law of motion where the net force causes the acceleration of an object. The force of gravity which is downward force and the normal force acting on an object which is perpendicular to the object are equal in magnitude but opposite to the direction, so while determining the net force, these two components cancel out. The net force is the horizontal component of the force and in our explanation, we consider everything as frictionless surface since friction should also be calculated while called the work-energy component of the object. The two most basics of energy classification are potential energy and kinetic energy. There are various kinds of kinetic energy like chemical, mechanical, thermal, nuclear, electrical, radiant energy, and so on. The work is done when there is a change in energy and it mainly depends on the application of force and movement of the object. Let us say how much work is needed to lift a 5kg ball 5m high. Work is mathematically represented as Force ×Displacement. So it will be 5kg times the gravitational constant on earth and the distance moved by the object. Wnet=Fnet times Displacement.
I included the previous parts for context, but I just need help on "Part D" please!
![**Learning Goal:**
To apply the law of conservation of energy to an object launched upward in Earth's gravitational field.
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 Earth. The energy transformations that take place involve the object's kinetic energy \( K = (1/2)mv^2 \) and its gravitational potential energy \( U = mgh \). 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:
\[ K_i + U_i = K_f + U_f \]
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.
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**First**, let us consider an object launched vertically upward with an initial speed \( v \). Neglect air resistance.
**Part A**
As the projectile goes upward, what energy changes take place?
- [ ] Both kinetic and potential energy decrease.
- [ ] Both kinetic and potential energy increase.
- [x] Kinetic energy decreases; potential energy increases.
- [ ] Kinetic energy increases; potential energy decreases.
**Correct.**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03d19364-d2f4-4703-a72b-e46c5aa833e9%2F5e85367b-e01c-42b4-a187-ceac2adedcc7%2F9sbiciw_processed.png&w=3840&q=75)
![## Part D
Using conservation of energy, find the maximum height \( h_{\text{max}} \) to which the object will rise.
Express your answer in terms of \( v \) and \( g \). You may or may not use all of these quantities.
### Input Box
- \( h_{\text{max}} = \) [Input Field]
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