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Mechanical Energy Pt. 2 Week
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Demonstrate Elastic Potential Energy
Objective Student will: Demonstrate Elastic Potential Energy Cornell Notes Needed – Section 11.1
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Elastic Potential Energy = ½ (spring constant)(distance of stretch)2
Cornell Notes (1/2) Questions How are elastic and gravitation PE the same? Elastic Potential Energy: Object’s energy that is stored in a stretched or compressed spring PEelastic = ½ kx2 Elastic Potential Energy = ½ (spring constant)(distance of stretch)2 Elastic Potential Energy - Joules (J) Spring Constant - Newton meter(Nm) Distance - meter(m) Spring constant is a measure of how hard it is to stretch or compress a spring and is dependent on how the construction of the spring Slinky Example
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Cornell Notes (2/2) The amount of energy stored in the spring depends on the distance the spring is stretched or compressed. What are two ways to store energy in a spring?
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Cornell Notes (1/5) Example: Elastic Potential Energy
Hawkeye pulls on an arrow 1 meter back and shoots it. The arrow reaches a velocity of m/s (173 mph). Assuming the bow has a spring constant of 150 N/m, what is the arrow’s mass? (Neglect air friction)
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Cornell Notes (2/5) Steps Define Given: x = -1 m v = m/s2 k = 150 N/m Unknown: m = ? Hawkeye pulls on an arrow 1 meter back and shoots it. The arrow reaches a velocity of m/s (173 mph). Assuming the bow has a spring constant of 150 N/m, what is the arrow’s mass? (Neglect air friction)
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Cornell Notes (3/5) Choose an equation or situation:
Plan Choose an equation or situation: Rearrange the equation to isolate the unknown: PE elastic = 1 2 k x 2 KE= 1 2 m v 2 PE elastic = 1 2 k x 2 KE= 1 2 m v 2 1 2 k x 2 = 1 2 m v 2 k x 2 =m v 2 k x 2 v 2 = m v 2 v 2 k x 2 v 2 =m
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Cornell Notes (4/5) k x 2 v 2 =m 150∙ −1 2 77.33 2 =m 1.93973878184=m
Calculate Substitute the values into the equation and solve k x 2 v 2 =m 150∙ − =m =m 1.940 kg=m
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The mass of the arrow turns out to be 1.940 kg or 4.277 lbs.
Cornell Notes (5/5) Evaluate The mass of the arrow turns out to be kg or lbs.
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