# Energy expended in acceleration

• Sep 12th 2013, 04:36 PM
Jock
Energy expended in acceleration
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.
• Sep 13th 2013, 05:49 AM
MINOANMAN
Re: Energy expended in acceleration
Show us some of your work...first...
• Sep 13th 2013, 05:53 AM
HallsofIvy
Re: Energy expended in acceleration
Quote:

Originally Posted by Jock
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

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.

Quote:

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.
• Sep 13th 2013, 06:57 AM
Jock
Re: Energy expended in acceleration
so how do you suggest I go about this? There are so many different formulas and really lost my way with this....
• Sep 13th 2013, 07:27 AM
HallsofIvy
Re: Energy expended in acceleration
I said "You need to use "v= at" and "d= (1/2)at^2" for velocity and distance at constant acceleration." Try that.
• Sep 13th 2013, 07:29 AM
Jock
Re: Energy expended in acceleration
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)????
• Sep 13th 2013, 04:31 PM
topsquark
Re: Energy expended in acceleration
Quote:

Originally Posted by Jock
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)????

They look okay. One point: I got $\displaystyle \mu _k = 0.1550$. (Don't round so much!) And remember you still need to do the graphs.

-Dan
• Sep 14th 2013, 06:03 AM
Jock
Re: Energy expended in acceleration
Yeah don't know how to get the figures for the graphs know what they should look like but just don't know the values