Help! Urgent: Conservation of Energy Review Questions

To apply this to the problem, one must use the relevant equation of motion (for constant acceleration) to determine the distance over which the friction force acts.
  • #1
fatalerror
2
0
These were review problems for the test I have tomorrow. My teacher was not present the day these were given, so I could not ask him for help. The answers were given for these problems, but in terms of getting to these answers, I am totally lost. Can someone please explain to me how to achieve the answers to these questions (answers are below questions in my explanations)? Thanks for any assistance. If you can only do one, then that's fine. Any explanations will do because I need to study with them for the test tomorrow afternoon. Again, thanks a bunch.
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1. A 0.280 kg block on a vertical spring with a spring constant of 5.00 x 10^3 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

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Wouldn't this have to do with elastic energy, so 1/2KX^2? The answer is 11m, but I don't know what method to use to achieve this answer.

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2. A 80 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 6.0 m/s.

How fast is he going as he lands on the trampoline, 2 m below?

If the trampoline behaves like a spring of spring constant 5.2 × 104 N/m, how far does he depress it?

(Ignore small changes in gravitational potential energy.)
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For the first part, I thought you would do: PE + KE (Before) = PE + KE (After). KE (before) and PE (after) would be 0 because he doesn't have any kinetic energy before he jumps, and he doesn't have any potential energy after he jumps. The answer is 8.67m/s, and again, I don't know what method to use to get the answer.

For the second sub-question, I thought you would again use the elastic energy equation, but alas I was wrong. The answer for the second question is 0.34m. I just don't know how to get this answer.

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3. A 14 kg child descends a slide 2.50 m high and reaches the bottom with a speed of 2.46 m/s. How much thermal energy due to friction was generated in this process?
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I would assume PE=KE though. The answer is 301 J.
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4. A 68 kg person escapes from a burning building by jumping from a window 25 m above a catching net. Assuming that air resistance exerts a 95 N force on the person during the fall, determine the person's velocity just before hitting the net.
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I would assume PE=KE though. The answer is -20.5 m/s.
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5. A roller coaster, shown in the figure (ha = 50 m, hb = 28 m, hc = 12 m), is pulled up to point 1 where it and its screaming occupants are released from rest. Assuming no friction, calculate the speed at points 2, 3, 4. The roller coaster cart is 500kg.

http://i26.tinypic.com/hurosi.gif - Link to figure.
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Totally lost on this one. The answers are:

31.3
m/s (at point 2)


20.8
m/s (at point 3)


27.3
m/s (at point 4)


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  • #2
fatalerror said:
These were review problems for the test I have tomorrow. My teacher was not present the day these were given, so I could not ask him for help. The answers were given for these problems, but in terms of getting to these answers, I am totally lost. Can someone please explain to me how to achieve the answers to these questions (answers are below questions in my explanations)? Thanks for any assistance. If you can only do one, then that's fine. Any explanations will do because I need to study with them for the test tomorrow afternoon. Again, thanks a bunch.
___________________________________________________________________1. A 0.280 kg block on a vertical spring with a spring constant of 5.00 x 10^3 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

_________________________________________________

Wouldn't this have to do with elastic energy, so 1/2KX^2? The answer is 11m, but I don't know what method to use to achieve this answer.
Assuming no energy losses
All the elastic potential energy is converted to gravitational potential energy.
You know that the elastic potential energy=[itex]\frac{1}{2}kx^2[/itex] and that should be the same as the gravitational potential energy. Do you know an equation to find the gravitational potential energy of an object of mass,m, at a height,h, in a gravitational of field of field strength,g?
fatalerror said:
3. A 14 kg child descends a slide 2.50 m high and reaches the bottom with a speed of 2.46 m/s. How much thermal energy due to friction was generated in this process?
__________________________________________________ ________________
I would assume PE=KE though. The answer is 301 J.
__________________________________________________ __________________
All the gravitational pe is converted into energy to cause the child to move at 2.46m/s and energy in overcoming friction.
Same concept for Q4

For Q5. If you are ignoring frictional effects. From point 1 to point 2, there is a change in height so the gravitational pe would change. What do you think all this energy is converted into?
 
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  • #3
1. A 0.280 kg block on a vertical spring with a spring constant of 5.00 x 10^3 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
There are three energies involved here - 1. spring mechanical (potential) energy, 2. gravitational potential energy, and 3. kinetic energy.

When released, the spring energy (1/2 kx2, where x = 0.110 m) is changes the gravitational potential energy (mgh) and kinetic energy of the block over the 0.110 m. So equate those. Then from the KE of the mass (1/2 mv2), that gives the initial velocity for the second part of the problem, which is a vertical launch.

See this for reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra1


for work done by (or lost to friction), one applied the definition of work = force * distance.
 
  • #4
Astronuc; if we do not know the height nor the velocity, how do we solve the equation?

Is it 1/2kx^2+mgh=1/2mv^2 for problem 1, part one?
 
  • #5
fatalerror said:
Astronuc; if we do not know the height nor the velocity, how do we solve the equation?

Hi!
Velocity at top most point will be zero:smile:
 

FAQ: Help! Urgent: Conservation of Energy Review Questions

What is conservation of energy?

Conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

Why is conservation of energy important?

Conservation of energy is important because it is a fundamental law of the universe that governs the behavior of energy and matter. Understanding and applying this principle allows us to make accurate predictions about the behavior of physical systems and develop efficient technologies.

How is conservation of energy related to the environment?

Conservation of energy is closely related to the environment because it helps us understand how energy is transferred and transformed within natural systems. This understanding is essential for managing and conserving natural resources, as well as developing sustainable energy sources.

What are some real-world examples of conservation of energy?

Some real-world examples of conservation of energy include the use of renewable energy sources like solar and wind power, which harness energy from the environment without depleting natural resources. Additionally, energy-efficient technologies such as LED light bulbs and hybrid cars also demonstrate the principles of conservation of energy.

How can we apply the principles of conservation of energy in our daily lives?

We can apply the principles of conservation of energy in our daily lives by practicing energy conservation habits, such as turning off lights and unplugging electronics when not in use, using public transportation or carpooling, and choosing energy-efficient products. These actions not only help to conserve energy and reduce our carbon footprint, but also save money on utility bills.

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