Part 5 of 5 - Hitting the Ground Next, calculate the same values as the box slides along the top of the hill. These values are presented in the same order in each part, but you may need to find them in a different order. (a) What is the kinetic energy of the box at the starting point? K2=1531.64 J (b) What is the gravitational potential energy of the box at the starting point? U₂=0J (c) What is the total energy of the box at the starting point? E2 1531.64 J II. Velocity Now, use this to find the velocity of the box at the top of the hill. (a) What is the velocity of the box at the top of the hill? W 11.8 m/s 11.8 m/s (0.5 × ŷ + x +0.5 × 2) 2
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.
![Sliding Up a Hill
When a problem contains only conservative forces, it is possible to model much of the dynamics using only energy conservation. In this application, you will return to the sliding box problem, but the ramp is replaced with a
strangely-shaped but frictionless slope. This makes this into a very difficult problem to solve kinematically, but straight-forward using energy conservation.
ū
h
You push a box of mass 22 kg with your car up to an icy hill slope of irregular shape to a height 5.2 m and width 4.5 m. The box has a speed 11.8 m/s when it starts up the hill, the same time that you brake. It then rises up to
the top (with no friction), which flattens off levelly for a distance 5 m. It ends at a sheer cliff, and then falls to the ground at the same level as where the box started (with no drag).
Use the ground as the zero for potential energies.
Part 5 of 5 - Hitting the Ground
Next, calculate the same values as the box slides along the top of the hill. These values are presented in the same order in each part, but you may need to find them in a different order. (a) What is the kinetic energy of the box
at the starting point?
K2=1531.64 J
(b) What is the gravitational potential energy of the box at the starting point?
U₂ =0J
(c) What is the total energy of the box at the starting point?
E2 = 1531.64 J
II. Velocity
Now, use this to find the velocity of the box at the top of the hill.
(a) What is the velocity of the box at the top of the hill?
-11.8 m/s
11.8 m/s
(0.5
128 Ý +0.5
× 2)
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