# Thread: Need some help with this question with energy

1. ## 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.

2. ## 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 $\displaystyle d = \frac 1 2 at^2$, solve for a. Then calculate velocity as a function of time: $\displaystyle v = at$. Finally KE as a function of time is $\displaystyle KE= \frac 1 2 mv^2 = \frac 1 2 m(at)^2$.

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

d) Use $\displaystyle \sum F = ma$. 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.

3. ## 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?

4. ## Re: Need some help with this question with energy

can anyone help me if im on right track with this

5. ## Re: Need some help with this question with energy

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

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.

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"?

Does this look somewhere in ball park?

6. ## Re: Need some help with this question with energy

This looks like something from teeside university

7. ## Re: Need some help with this question with energy

it is are you doing the same course