Show us some of your work...first...
A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal
surface. The mass is initially at rest and covers a distance of 5 m in 0.92 s
under the action of the force. Assuming there are no energy losses due to
air resistance and therefore that the acceleration is constant:
(a) Calculate the total energy expended in the acceleration.
(b) Plot a graph of the kinetic energy of the mass against time.
(c) Plot a graph of the kinetic energy of the mass against distance.
(d) Calculate the coefficient of friction between the mass and the surface. Any help with this would be great. Only got bits for (a) velocity = 5.435 acceleration = 5.908 Kinetic energy = 104.71 frictional force = 58.86 co-efficient of friction = 1.36 and so far total energy expended = 504.96 J. Like I said any help would be great really need it.
This is wrong to begin with. If an object were moving at constant velocity 5.435 m/s, it would cover 5 meters in 0.92 seconds but this is NOT at constant velocity, it is at constant acceleration. You need to use "v= at" and "d= (1/2)at^2" for velocity and distance at constant acceleration.
acceleration = 5.908 Kinetic energy = 104.71 frictional force = 58.86 co-efficient of friction = 1.36 and so far total energy expended = 504.96 J. Like I said any help would be great really need it.
Did somemore number crunching and have came up with the following. Lets see if this is right feed back would be great. Work done =fd= 400J Average velocity=5.438 accelearation= 10.87 Final kinetic energy (energy expended) = 354J Lost energy due to friction = 45.6J friction force = 9.11N therefore the coeffient of friction is 0.15 Would these answers work for (a) and (d)????