176 reputation
7
bio website
location Norway
age
visits member for 2 years, 2 months
seen Jul 9 '12 at 15:02

Pre-med


May
6
awarded  Self-Learner
Jun
4
awarded  Teacher
May
22
awarded  Citizen Patrol
May
22
revised From knowing just the change in kinetic energy, can we find the friction force and engine power?
added 105 characters in body
May
22
suggested suggested edit on From knowing just the change in kinetic energy, can we find the friction force and engine power?
May
22
comment From knowing just the change in kinetic energy, can we find the friction force and engine power?
What I meant was that 1500 is the total pull forward by the car and 1000N is the friction.
May
22
comment From knowing just the change in kinetic energy, can we find the friction force and engine power?
Lets say a car changes speed from 10 m/s to 20m/s. That's an increase in kinetic energy of 60kj. Let's assume weight of car 1200kg, and that the acceleration takes 8 seconds. delta v / delta t, gives 1.25m/s^2 as acceleration. Using road formula i get 120 meters. 60kj/12 = 500N. That would be the force that actually accelerates the car. But this isn't general, because I had to make up a few facts. If I would have chosen a different time the force would have been much greater. But one more detail, now!. 1200kg*1.25 = 1500N. 1500-500 = 1000N. Is the friction 1000N i this case?
May
21
awarded  Editor
May
21
revised From knowing just the change in kinetic energy, can we find the friction force and engine power?
added 107 characters in body
May
21
asked From knowing just the change in kinetic energy, can we find the friction force and engine power?
May
20
comment My book uses helium weight in reaction formula instead of weight alpha particle?
I get that it's a very small error to make. I thought accuracy was a good thing :(
May
20
asked My book uses helium weight in reaction formula instead of weight alpha particle?
May
19
awarded  Supporter
May
19
awarded  Scholar
May
19
accepted Atomic weight in respect to the binding energy?
May
19
asked Atomic weight in respect to the binding energy?
May
17
comment I have a slight problem understanding the concept of “work”?
It makes a lot more sense now! And if the force needed to lift the car at a constant speed is m*g, then if using the "work" rule you would still get the work with (m*g)*height which is the same. Thanks.
May
17
comment I have a slight problem understanding the concept of “work”?
Do you mean mgh?
May
17
comment I have a slight problem understanding the concept of “work”?
Specifically the question states that the work done on the car is 0 during the movement of the car and then it states that this is due to the kinetic energy being 0 at the start and 0 at the end. Is this wrong? In that case it's just that stupid question at the site that threw me off.
May
17
comment I have a slight problem understanding the concept of “work”?
This chapter assumes movement only along a road or up and down. So we never need the cosine part of the work formula. Even so the book clearly states that there is only under two circumstances that physical labor doesn't count as work; when the movement is 0, and when the force is 90 degrees in respect to direction of movement. The origin of my confusion was an online chapter test at the site of the book, that's where it said that the work done by the crane was 0. I will look long and hard at this today.