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Transcript
Energy of the universe is conserved
βˆ†πΈπ‘’π‘›π‘–π‘£ = 0
Mechanical Energy
πΈπ‘šπ‘’π‘β„Ž = π‘ˆ + 𝐾
Gravitational Potential Energy
π‘ˆπ‘”π‘Ÿπ‘Žπ‘£ = π‘šπ‘”β„Ž
Elastic Potential Energy
π‘ˆπ‘’π‘™π‘Žπ‘ π‘‘π‘–π‘
1
= π‘˜βˆ†π‘₯ 2
2
Translational Kinetic Energy
1
𝐾 = π‘šπ‘£ 2
2
Work
𝑑𝑓
π‘Š = ∫ 𝐹⃗ βˆ™ π‘‘π‘Ÿβƒ—
𝑑𝑖
So this is the dot product of force and displacement. The
definition above is strictly true for a time-varying force.
For a constant force, we can say:
π‘Š = 𝐹⃗ βˆ™ π‘Ÿβƒ— = πΉπ‘Ÿπ‘π‘œπ‘ (πœƒ)
(Note: implications for angle between force and
displacement)
We have the work-kinetic energy theorem:
π‘Šπ‘‘π‘œπ‘‘π‘Žπ‘™ = βˆ†πΎ
But we have different kinds of work depending on types of
forces
π‘Šπ‘‘π‘œπ‘‘π‘Žπ‘™ = π‘Šπ‘π‘œπ‘›π‘  + π‘Šπ‘›π‘œπ‘›π‘π‘œπ‘›π‘ 
Conservative forces (gravity/elastic in this class)
π‘Šπ‘π‘œπ‘›π‘  = βˆ’βˆ†π‘ˆ
Therefore, combining our work-kinetic energy theorem
and the two equations above:
π‘Šπ‘‘π‘œπ‘‘π‘Žπ‘™ = βˆ†πΎ = π‘Šπ‘π‘œπ‘›π‘  + π‘Šπ‘›π‘œπ‘›π‘π‘œπ‘›π‘ 
βˆ†πΎ = βˆ’βˆ†π‘ˆ + π‘Šπ‘›π‘œπ‘›π‘π‘œπ‘›π‘ 
π‘Šπ‘›π‘œπ‘›π‘π‘œπ‘›π‘  = βˆ†πΎ + βˆ†π‘ˆ = βˆ†πΈπ‘šπ‘’π‘β„Ž
1. The graphs above show the force a person exerts on a box sliding on a frictionless surface
without air drag with respect to both time and position.
a. Find the work done by the person on the box for the first 6 meters of pushing.
b. Find the change in momentum of the box from t = 0 seconds to t = 6 seconds.
2. A horizontal spring has an object attached to it as shown. At relaxed length, the object
moves with a speed of 5 m/s. Assume no friction/drag.
a. What is the maximum displacement of the object?
b. Graph total mechanical energy, potential energy, and kinetic energy as a function
of displacement.
c. How fast is the object moving when the spring is compressed 0.2m from its
relaxed length?
3. A ball (m = 3 kg) falls down a frictionless chute from an initial height of 10m, coming
out horizontally on frictionless ice. After sliding across the ground for a bit, the ball
encounters a rough patch of dirt (µk = 0.3) with a length of 4 meters. After the rough
patch, it rolls up a frictionless slope.
a. What is the speed of the ball as it emerges from the chute?
b. What is the speed of the ball just after it goes across the dirt?
c. What is the height of the ball at the top of the final slope?
4. A car (m = 1200 kg) skids to a stop in a distance of 50 meters from a speed of 40 m/s.
Calculate the work done by:
a. Gravity
b. Friction (µk = 0.7)
c. Air drag