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.

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.

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?

Re: Need some help with this question with energy

can anyone help me if im on right track with this

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

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?

Re: Need some help with this question with energy

This looks like something from teeside university

Re: Need some help with this question with energy

it is are you doing the same course