- #1
mark9159
- 13
- 0
greetings...i need some help please and someone to check if i did the problems correctly.
1.) How much kinetic energy will a 5kg ball have after rolling down a 1m high incline?
My Answer: 4.43 J
If Potential Energy=mgh, then P=(5kg)(9.81m/s^2)(1m) so P=49.05 J. If this is true, then Kinetic energy = 49.05 J = 1/2mv^2. 49.05=1/2(5)v^2. Then i divided 49.05 by 2.5 (49.05/2.5) which gave me v^2=19.62. I then took the square root of 19.62 which is approximately 4.43 J.
2.) Why are no collisions perfectly inelastic?
My Answer: Almost all collisions involve some rebounding.
3.) A lever is used to raise a 300N rock a distance of 0.6 meters. You must use a force of 75N to accomplish this task. What work is done?
My Answer: 2.45 J
If work can equal mass times gravity times height, then work equals 300N times 0.6 meters, which equals 180. I then did 180 divided by 75N (the amount of force needed), coming up with the solution: 2.45 J
thank you very much for checking my work!
mark
1.) How much kinetic energy will a 5kg ball have after rolling down a 1m high incline?
My Answer: 4.43 J
If Potential Energy=mgh, then P=(5kg)(9.81m/s^2)(1m) so P=49.05 J. If this is true, then Kinetic energy = 49.05 J = 1/2mv^2. 49.05=1/2(5)v^2. Then i divided 49.05 by 2.5 (49.05/2.5) which gave me v^2=19.62. I then took the square root of 19.62 which is approximately 4.43 J.
2.) Why are no collisions perfectly inelastic?
My Answer: Almost all collisions involve some rebounding.
3.) A lever is used to raise a 300N rock a distance of 0.6 meters. You must use a force of 75N to accomplish this task. What work is done?
My Answer: 2.45 J
If work can equal mass times gravity times height, then work equals 300N times 0.6 meters, which equals 180. I then did 180 divided by 75N (the amount of force needed), coming up with the solution: 2.45 J
thank you very much for checking my work!
mark