1. Aim: How do we apply conservation of energy to solving problems involving springs?

2. Do Now: (Yesterday’s Atwood’s Machine)
Answer each of the following:
What form(s) of energy does mass A possess initially? D
What form(s) of energy does mass B possess initially? B
What form(s) of energy does mass A possess in the final state of the system shown? C
What form(s) of energy does mass B possess in the final state of the system shown? A
Kinetic Energy
Gravitational Potential Energy
Both
None of the above

3. Calculating the Elastic Potential Energy in a Dropper
Think Pair and Share: How could we calculate the elastic potential energy in this dropper?
Calculate the initial and final gravitational potential energies of the toy. The difference between these values must be equal to the elastic potential energy in the toy.

4. Conservation of Mechanical Energy
Emec=KE +U In an isolated system, mechanical energy is conserved. The sum of kinetic and potential energy for any state of a system does not change over time.

5. Tips for solving Conservation of Mechanical Energy problems
Determine the initial and final energy states of your system. The type of energy present will change but the total mechanical energy of the system will stay the same.

6. Block-Spring Collision
A block of mass kg is given an initial velocity vo = 1.20 m/s to the right and collides with a light spring force of spring constant k=50 N/m.
Calculate the maximum compression of the spring after the collision.
Before we solve this, do the following:
In 1-2 complete sentences, write down the energy transformation that occurs during this collision.
Express the initial and final energy states.
Use conservation of energy to solve for the unknown compression.
0.152 m

7. Block-Spring Collision
The kinetic energy transforms into elastic potential energy.
Ei=Ef
1/2mv2=1/2kx m=0.8 kg, v =1.2 m/s, k=50 N/m
X=0.15 m

8. Dropping a Block onto a Spring
A spring that has a force constant of 1000 N/m is placed on a table in a vertical position. A block of mass 1.60 kg is held 1.00 m above the free end of the spring. The block is dropped from rest so that it falls vertically onto the spring. By what distance does the spring compress?
d=0.19 m

9. Dropping Block onto Spring
Before we solve this,
In 1-2 complete sentences, write down the energy transformation that occurs during this collision.
Express the initial and final energy states.
Use conservation of energy to solve for the unknown compression.
The gravitational potential energy transforms into elastic potential energy
Ei = Ef
mg(1+x) = 1/2kx2
1.6(10)(1+x)=1/2(1000)x2 solve for x using a quadratic formula and get x =0.2 m

10. A block pushing a Spring
A 3 kg block on a frictionless horizontal surface is pushed against a spring that has a spring constant of 400 N/m, compressing a spring by 0.30 m. The block is released, and the spring projects it along the surface and then up a frictionless incline of angle 37 degrees.
Find the speed of the block just after it leaves the spring?
How high does the block go?
Before, we solve this, explain in 1-2 sentences how the type of energy that the block possesses changes from when it is launched by the spring to when it goes up the hill.
a) 3.16 m/s b) 0.72 m

11. Block Pushing a Spring
The elastic potential energy is converted into kinetic energy as the box leaves the spring and then the kinetic energy is converted into gravitational potential energy as the box goes up the hill
1/2kx2 = 1/2mv2
½(400)(.3)2=1/2(3)v2
v=3.46 m/s
b) 1/2kx2=mgh
½(400)(.3)2=(3)(10)h
H=0.6 m

12. Group Activity: Analyze the Bungee System
A 61 kg bungee-cord jumper is on a bridge 45 m above a river. The elastic bungee has a relaxed length of L=25.0 m. Assume the cord has a spring constant of 160 N/m. If the jumper stops before reaching the water, what is the height h of her feet above the water at her lowest point?

13. Bungee Jumper
The gravitational potential energy is converted into elastic potential energy

14. Solution
Ei=Ef Mg(L+d)=1/2k(d)2 61(10)(25+d)=1/2(160)d d=80d2 0=80d2-610d solve a quadratic formula for d and get d= m Since the bridge is 45 m above the river so h= 45-(L+d)=45-( )=1.86 m h=1.86 m