0.21 Phy1190: energy -- kinetic and mechanical energy  (Page 8/8)

Answers:

1. Assuming that the dart was not deformed as it was forced out of the barrel of the gun, the potential energy of the dart is given by

mass*gravity*height

Enter the following into the Google search box:

(0.1 kg) * (9.8 (m / (s^2))) * (1 m)

The result should be:

Potential energy = 0.98 joules

2. The kinetic energy of the dart is given by

0.5*m*v^2

Enter the following into the Google search box:

0.5*0.1kg*(10m/s)^2

The result should be:

kinetic energy = 5 joules

3. The total mechanical energy of the dart is the simple sum of the potential energy and the kinetic energy, which is:

Total mechanical energy = 5.98 joules

4. When the dart exits the gun, the gravitational potential energy of the gun is reduced because the total mass of the gun and the dart is reduced by the massof the dart.

In addition, the elastic potential energy stored in the spring is imparted into the dart in the form of kinetic energy.

Therefore, the loss in mechanical energy of the gun is equal to the total mechanical energy of the dart immediately upon exit from the gun barrel.Therefore, the total mechanical energy of the gun is reduced by 5.98 joules when the dart exits the gun.

A crate on a ski run

A crate containing soft drinks with a mass of 5 kg is accidentally released at the top of a ski run and slides down the ski run to the valley below. Theheight of the point where the crate is released is 100 m above the valley floor. The crate goes through numerous dips and over many small hills on the way down but never stops.

Assuming there is no friction, no air resistance, no deformation, and no loss of energy in any form during the trip, what is the magnitude of the crate's velocitywhen it reaches the valley floor?

Answer:

As presented, this is a simple case of the conversion of gravitational potential energy into kinetic energy. The fact that the crate slowed down andsped up several times during the trip while negotiating little dips and hills doesn't matter. All that really matters is the balance of energy between the starting point and point where the cratereached the valley floor. With no energy loss during the trip, the total mechanical energy at the end of the trip must equal the total mechanical energy at thebeginning of the trip.

At the top of the hill, the crate's gravitational potential energy was equal to

m*g*h = 5kg*(9.8m/s^2)*100m = 4900 joules

Therefore, at the bottom of the hill, with no remaining potential energy, the crate's kinetic energy must be equal to

0.5*m*v^2 = 4900 joules

Rearranging terms gives us

v^2 = (4900 joules)/(0.5*m), or

v = sqrt((4900 joules)/(0.5*m)), or

v = sqrt((4900 joules)/(0.5*5kg))

Entering this expression into the Google calculator gives us the crate's velocity when it reached the valley floor as

v = 44.3 m/s

Do the calculations

I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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