1. ok m g h = 20 * 10 * 2 Joules
2. No. That 400 Joules of potential energy becomes 400 Joules of kinetic energy during the fall.
3. Yes - At maximum stretch the ball stops. The ball is now 2 meters plus x below the starting point where x is the stretch distance down. The total energy is all potential in the springs and gravitational and is m g (2 + x)
4. Energy is conserved I suppose.
Use g=10 N/kg in this assignment.
1. What is the gravitational energy (relative to the unstretched surface of the trampoline) of the 20 kg ball at its apex 2 m above the trampoline?
I got 400 J
2. What is the kinetic energy of the ball just before impacting the trampoline?
I got 0 J
3. At maximum stretch at the bottom of the motion, what is the sum of the elastic and gravitational energy of the ball?
Is it because the sum of elastic and potential energy is equal to the change in gravitational energy from original position?
4. What conclusions can be drawn from the answers above
I'm not too sure what to put as my answer
2 answers
1.What is the gravitational energy (relative to the unstretched surface of the trampoline) of the 20kg ball at its apex 2.0m above the trampoline?
KE = GPE= mgh
m g h = 20 x 10 x 2 Joules= 400 J
The kinetic energy just before it hits the trampoline is equal to the gravitational potential energy at its apex.
2.What is the kinetic energy of the ball just before impacting the trampoline
All energy is transferred into kinetic energy therefore Ek = 400J
3. At maximum stretch at the bottom of the motion, what is the sum of elastic and gravitational energy of the ball?
If x is the extension in the spring, then h1 = h2 + x + y.
4. What conclusions can you draw for the answers above
Through the answers above, I can conclude that energy is always conserved through the law of conservation of energy
Sum of gravitational and elastic energy at maximums stretch.
height=2 meters+x
where x is the stretch distance
Sum=mg(2+x) J
Simplifying the above expression, it becomes
sum=20×10(2+x)
sum=400+220x J
Question 4
Conclusion made: Energy is conserved during the process.
Question 5
Equivalence in the energy is caused by difference in mass and height between the ball and the wrestler since the energy is a function of mass, gravity and height.
Question 8
To maintain the same height in every bounce, a person must apply enough force to return to the same height as the previous bounce.
KE = GPE= mgh
m g h = 20 x 10 x 2 Joules= 400 J
The kinetic energy just before it hits the trampoline is equal to the gravitational potential energy at its apex.
2.What is the kinetic energy of the ball just before impacting the trampoline
All energy is transferred into kinetic energy therefore Ek = 400J
3. At maximum stretch at the bottom of the motion, what is the sum of elastic and gravitational energy of the ball?
If x is the extension in the spring, then h1 = h2 + x + y.
4. What conclusions can you draw for the answers above
Through the answers above, I can conclude that energy is always conserved through the law of conservation of energy
Sum of gravitational and elastic energy at maximums stretch.
height=2 meters+x
where x is the stretch distance
Sum=mg(2+x) J
Simplifying the above expression, it becomes
sum=20×10(2+x)
sum=400+220x J
Question 4
Conclusion made: Energy is conserved during the process.
Question 5
Equivalence in the energy is caused by difference in mass and height between the ball and the wrestler since the energy is a function of mass, gravity and height.
Question 8
To maintain the same height in every bounce, a person must apply enough force to return to the same height as the previous bounce.
1 answer