albertlee Posted April 13, 2004 Share Posted April 13, 2004 How to find out the power by using average velocity? given info: object:1kg, average velocity:1m/s, distance:10m Aprciate for the responds Link to comment Share on other sites More sharing options...
wolfson Posted April 13, 2004 Share Posted April 13, 2004 Kinetic energy = 1/2 mv^2 m= mass v= velocity Units of Joule PE = mgh m= mass g= gravity (9.81ms^-2) h = height (metres) Unity joule Average power = workdone / time (secs) Work done = force (9.81) * distance moved. If a force F acting on a particle has accelerated the particle to a velocity v , the instantaneous power expended is: power = F * v There's a few off the top of my head. Link to comment Share on other sites More sharing options...
darkbob5150 Posted April 13, 2004 Share Posted April 13, 2004 How to find out the power by using average velocity? given info: object:1kg' date=' average velocity:1m/s, distance:10m [/quote'] I take it you mean frictionless horizontal motion so no gravity issues. Power is the rate of doing work - and is also given by Force x (average) Velocity. What I'd do is work out what constant acceleration over a distance of 10m you need to produce an average velocity of 1m/s. Average velocity over a period of constant acceleration is given by (Final velocity - initial velocity)/2. In this case (assuming you start from rest) the final velocity is 2m/s and the time taken to travel the 10m would be distance/average velocity = 10sec. So the acceleration (change in velocity with time) would be 0.2 m/s^2 You now have the force you would have to apply to produce this acceleration (assuming frictionless horizontal motion) from F = ma. Put the numbers into Power = Force x average velocity and that'll be it. Work is given by Force x Distance moved (in the direction of the Force) so if you wanted to do it that way then you have to divide by the time to get the Power. Link to comment Share on other sites More sharing options...
albertlee Posted April 13, 2004 Author Share Posted April 13, 2004 Thx Dark........But the equation from Wolfson confused me: k.e. = 1/2mv^2 since m=1kg, v=av*2=2m/s Can I work it as k.e.=1/2(1)(2^2)? But what I wonder is where does this equation take the distance into acount? Any help? thx Link to comment Share on other sites More sharing options...
Dave Posted April 13, 2004 Share Posted April 13, 2004 It doesn't, it just takes into account the kinetic energy an object has at a specific instant in time. Link to comment Share on other sites More sharing options...
albertlee Posted April 13, 2004 Author Share Posted April 13, 2004 Ok, just for sure, is my way of doing correct in my previous message? Link to comment Share on other sites More sharing options...
darkbob5150 Posted April 13, 2004 Share Posted April 13, 2004 Thx Dark........But the equation from Wolfson confused me: k.e. = 1/2mv^2since m=1kg' date=' v=av*2=2m/s Can I work it as k.e.=1/2(1)(2^2)? But [b']what I wonder is where does this equation take the distance into acount?[/b] Any help? thx Ok, the equation for kinetic energy - as Dave said - gives you the energy that the object has, at a particular time, because it is moving. You would use the KE in a question when there are no external force acting because energy would then be a conserved quantity - the amount of energy (kinetic + potential) would remain constant allowing you to work nice little things out. Because this question is about average velocities it means that the velocity changes (accelerates) as the object moves which would indicate that there are external (unbalenced) forces acting. External forces mean that you can't apply the conservation of energy because the force will either add energy to the system (your moving object) like pushing it to make it go faster, or take it away like friction. If you take the extra energy that the force introduces to the system into account then you can use the energy formula: Work done on an object is really just the change in energy of that object due to the applied force. It only depends on the initial and final states which is a useful and important feature. In this case (if you take the setup I had earlier where the object starts at rest, constantly accelerates and reaches a top speed of 2m/s after 10m) you can take the work done on the object as the change in KE ( = Final - Initial = 1/2 x m x 2^2 - 0 ) and you get the same answer as before for the work. Divide by the time taken to get the power. Link to comment Share on other sites More sharing options...
darkbob5150 Posted April 13, 2004 Share Posted April 13, 2004 sorry, just read the numbers in your reply properly. Yeah, you had it right - didn't need all that boring crap I wrote, but it explains a little bit as to why it's right! Soz -1 Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Thx Dark, the information is relavant and helpful......... Anyway, I have also problems with the direction of forces................. Can anyone tell me the basic concepts of the direction of force? for eg, how can an object move diagonally? or move in an indefined route, since it makes even harder to measure the distance and to say the direction of the force......... Aprciate for furthur helps! Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Anybody? Link to comment Share on other sites More sharing options...
swansont Posted April 14, 2004 Share Posted April 14, 2004 Thx Dark' date=' the information is relavant and helpful......... Anyway, I have also problems with the direction of forces................. [b']Can anyone tell me the basic concepts of the direction of force?[/b] for eg, how can an object move diagonally? or move in an indefined route, since it makes even harder to measure the distance and to say the direction of the force......... Aprciate for furthur helps! The net force is always in the direction of the acceleration, but this doesn't mean that it has to be in the direction of the displacement (which is given by the velocity vector) "Diagonal" motion is just an artifact of a coordinate system, which is arbitrary. 1 Link to comment Share on other sites More sharing options...
darkbob5150 Posted April 14, 2004 Share Posted April 14, 2004 Sounds like you need Newton's first law of motion: A body will continue in a state of rest or uniform motion in a straight line unless acted on by an unbalanced external force. When an unbalanced force acts on a body it will cause it to accelerate in the direction of the applied force. The acceleration lasts for as long as the force is applied. [The important thing about forces are that they are vectors, that means that every force must have both a magnitude (how hard you push it) and a direction (what direction you push it in).] So forces tend to change the direction something travels in and/or change the objects speed. I'm not too sure what you mean by your questions......an object travelling in a diagonal line is not under the influence of an unbalanced force (unless it's speed is changing). Do you mean it's moving under gravity? An undefined force that is changing it's magnitude / direction would cause the object to change it's speed / direction accordingly and travel in an undefined path. Here's an example that might help...? If you throw a ball diagonally up into the air (like you were throwing it to someone a few metres away) then think about what that would do. The gravitational force acts straight down with a magnitude of: Mass of the ball x Acceleration due to Gravity (from Newton's 2nd law F = Ma). So the vertical (up and down) motion of the ball is subject to an acceleration and will therefore cause the ball to slow down when heading for the sky, stop and then speed up towards the ground. The force due to gravity is constant, so the speed of the ball is always changing by a constant amount. The horizontal motion has no gravitational force acting on it, so the ball's direction / speed parallel to the ground does not change - your friend catches it with the same horizontal speed as you threw it with. This technique (resolving components of motion) allows you to resolve an objects motion into directions at right angles to each other to make a problem simpler. Here you can make the horizontal motion separate to the vertical motion so the horizontal motion is just a constant motion in a straight line problem, and the vertical motion is just acceleration in one dimension. Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 ok,maybe my question is not complete enough........How can explain the direction of force of a plane flying diagonally? if the direction of force is horizontal? Please help Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 If the previous message is not clear enough, here is an example which i confuse: How can we find the work done? according to direction of force and the distance? Please help! apreciate Link to comment Share on other sites More sharing options...
wolfson Posted April 14, 2004 Share Posted April 14, 2004 Work = F * Cos degrees Thus: Work = (400 N) * (30m) * (Cos 30) = 10392.30 J = 10.39 KJ Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Wolfson, What is Cos degree? and How does it work as a concept to the direction of the force? Please help. Link to comment Share on other sites More sharing options...
wolfson Posted April 14, 2004 Share Posted April 14, 2004 You need to use |cos| to calculate the force, it is used as you have a angle, and the only way we can calculate 400N on a 30 degress elavation is by using cosine, (you may also need to use |sine| in other parts of physics), have you come accross trigonometry?, same concept here. So by using cos (seen an your calculator as cos), we can use Equation 1. Work = F * Cos theta EQUATION 1 thus We can find out the work done (SI unit joules) do you understand? Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Ok, wolfson, the key to this whole question is the problem of cos..... First of all, I know nothing about "cos"...And i dont know what it is used for, and how,,, The only thing I know is that i can see a cos button on my calculator....... Any futhur help on cos? Link to comment Share on other sites More sharing options...
swansont Posted April 14, 2004 Share Posted April 14, 2004 You need to break the forces down into their x and y components. Only the net force in the x direction (horizontal) does any work for displacement in that direction. Any components in the y direction make no contribution. Link to comment Share on other sites More sharing options...
wolfson Posted April 14, 2004 Share Posted April 14, 2004 You have not done trig so just imagine it as if you have a degree on the horizontal you NEED to use cosine. Link to comment Share on other sites More sharing options...
wolfson Posted April 14, 2004 Share Posted April 14, 2004 Or i could teach you trig. Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Ok....Got a bit of sense.........it sounds like vectorial analysis.... Anyway, I am still very fresh with cosine....... Any body helps? Link to comment Share on other sites More sharing options...
albertlee Posted April 14, 2004 Author Share Posted April 14, 2004 Oh, forget about my previous message........... I know the concept of cosine just now...............it is adj./hyp. in a right-angled triangle, but....... What i really confused is still about how does it do to the direction of the forces and distance. Any help? Link to comment Share on other sites More sharing options...
albertlee Posted April 15, 2004 Author Share Posted April 15, 2004 Any body? Link to comment Share on other sites More sharing options...
wolfson Posted April 15, 2004 Share Posted April 15, 2004 it is adj./hyp. in a right-angled triangle So you have done trig Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now