1) State the Law of Conservation of Mechanical Energy. 2) A 5000-kg car is initi
ID: 1488482 • Letter: 1
Question
1) State the Law of Conservation of Mechanical Energy.
2) A 5000-kg car is initially coasting at a speed of 30 m/s when it comes to a Ø = 15.0° incline. It coasts without friction or air resistance, up the incline.
a) How large is the initial kinetic energy of the car, in kJ?
b) After it travels 85.0 m along the incline, how large is the gravitational potential energy of the car, in kJ? Take the zero of PE to be at the bottom of the incline?
c) How fast is the car coasting now after it went 70.0 m along the incline?
Explanation / Answer
1.
Conservation of mechanical energy states that the mechanical energy of an isolated system remains constant in time, as long as the system is free of all frictional forces. In any real situation, frictional forces and other non-conservative forces are always present, but in many cases their effects on the system are so small that the principle of conservation of mechanical energy can be used as a fair approximation. An example of a such a system is shown in . Though energy cannot be created nor destroyed in an isolated system, it can be internally converted to any other form of energy
An example of a mechanical system: A satellite is orbiting the Earth only influenced by the conservative gravitational force and the mechanical energy is therefore conserved. This acceleration is represented by a green acceleration vector and the velocity is represented by a red velocity vector
2 solution
A) The kinetic energy (Ekin) is defended by this formula Ekin=1/2*m*v^2.
So the Ekin is =1/2 * 5000kg * 30^2 = 2250000J = 2250 kJ.
B) The height above the zero PE level is = sin(15°) * 85m=22 m
The potential energy is defined as =m*g*h
So the potential energy is =5000kg * 9.8m/s^2 * 22 m=1078000J = 1078 kJ
C)the energy needs to be conserved, this means 0.5*m*v0^2=m*g*h+0.5*m*v1^2
Now it's easier to plug in than to solve first for v(1):
2250000J = 1078000 J+0.5*m*v1^2
1172000 = 0.5*m*v1^2
1172000 = 2500kg*v12
v1 = 21.65 m/s
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