# Need some help with this question with energy

• February 13th 2013, 05:57 AM
gordonmckee
Need some help with this question with energy
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.
• February 13th 2013, 08:56 AM
ebaines
Re: Need some help with this question with energy
a) Remember that energy expended = work = force applied times distance.

b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from $d = \frac 1 2 at^2$, solve for a. Then calculate velocity as a function of time: $v = at$. Finally KE as a function of time is $KE= \frac 1 2 mv^2 = \frac 1 2 m(at)^2$.

c) for KE as a function of distance use the formula $v^2 = 2ad$, so $KE = \frac 1 2 mv^2 = \frac 1 2 m(2ad)$.

d) Use $\sum F = ma$. From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
• February 21st 2013, 09:36 AM
gordonmckee
Re: Need some help with this question with energy
Thanks for that

Q1/
a) Remember that energy expended = work = force applied times distance.
80 X 5
400N

b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from , solve for a. Then calculate velocity as a function of time: . Finally KE as a function of time is .
A=dt2 / (1/2) => 9.2ms -2 then
V = at => 9.2 x 0.92 = 8.64 then
KE = ½ mv2 => ½ m (at)2 => ½ x 6 x (9.2 x 0.92) 2 => 214.92J

c) for KE as a function of distance use the formula , so .
KE = ½ mv2 = ½ m (2ad) = ½ x 6 x (2 x 9.2 x 5) = 270J

d) Use . From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
F = uRn Assume g=9.81ms-2 Rn = w = mg = 6 x 9.81 = 58.86N
P = F = 80N therefore P =uRn therefore u=P / Rn = 80 / 58.86 =1.35916 is the coefficient

Does this look somewhere in ball park?
• March 15th 2013, 12:14 PM
gordonmckee
Re: Need some help with this question with energy
can anyone help me if im on right track with this
• March 15th 2013, 02:00 PM
HallsofIvy
Re: Need some help with this question with energy
Quote:

Originally Posted by gordonmckee
Thanks for that

Q1/
a) Remember that energy expended = work = force applied times distance.
80 X 5
400N

b) Calculate the mass's acceleration from the data that it moves 5 m in 0.92 seconds: from , solve for a. Then calculate velocity as a function of time: . Finally KE as a function of time is .
A=dt2 / (1/2) => 9.2ms -2 then
V = at => 9.2 x 0.92 = 8.64 then
KE = ½ mv2 => ½ m (at)2 => ½ x 6 x (9.2 x 0.92) 2 => 214.92J

That's not what ebaines said and it is not even clear what question you are answering here. The question asked you to "Plot a graph of the kinetic energy of the mass against time." I can understand you not posting a graph here but you have not even calculated KE as a function of time.

"from $d= \frac{1}{2}at^2$ , solve for a" That gives $a= \frac{2d}{t^2}$, NOT " $a= at^2/(1/2)$". a is NOT 9.2.

Quote:

c) for KE as a function of distance use the formula , so.
KE = ½ mv2 = ½ m (2ad) = ½ x 6 x (2 x 9.2 x 5) = 270J
That's just bad arithmetic! ½ x 6 x (2 x 9.2 x 5)= 30 x 9.2= 276 J.
But, in any case, you were asked to "Plot a graph of the kinetic energy of the mass against distance." so you need KE as a function of d, not a single number:
KE= (1/2)m(2ad)= (6)(a)d. As I said before, your value for a is incorrect.

Quote:

d) Use . From this find the sum of forces. There are two forces acting: the 80N push and the resisting force of friction. So force of friction = pushing force - ma.
F = uRn Assume g=9.81ms-2

Rn = w = mg = 6 x 9.81 = 58.86N
P = F = 80N therefore P =uRn therefore u=P / Rn = 80 / 58.86 =1.35916 is the coefficient
What happened to the "-ma"?

Quote:

Does this look somewhere in ball park?
• July 3rd 2013, 10:19 PM
damon669
Re: Need some help with this question with energy
This looks like something from teeside university
• March 25th 2014, 10:16 PM
gordonmckee
Re: Need some help with this question with energy
it is are you doing the same course