Mechanics, energy, acceleration.

[oxon88]

Homework Statement

A horizontal force of 80N acts on a mass of 6Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

Homework Equations

v = s/t

a = (v-u) / t

s = ut + (1/2)a.t²

F = m.a

Ke = 1/2*m*V^2

The Attempt at a Solution

first I calculated the acceleration:

s = (1/2).a.t²

5 = (1/2)a.0.92²

10/0.92² = a = 11.815 ms^-2Then i calculated the net force:

F = m.a

F = 6 * 11.815 = 70.89N

therefore i think friction = 80N - 70.89N = 9.11N Would this be the coefficient of friction?Next i calculated the speed after 0.92s. s = d/t = 5/0.92 = 5.435 ms^-1

Then calculate kinetic energy: = (1/2)m.v² = 3*5.435² = 88.62Jso now would the total energy expended = kinetic energy + Friction force * 5m?

= 88.62 + 9.11 * 5 = 488.65 Nm
can anyone verify this? and let me know if I'm way off please?

 

[PeterO]

Homework Statement

A horizontal force of 80N acts on a mass of 6Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

Homework Equations

v = s/t

a = (v-u) / t

s = ut + (1/2)a.t²

F = m.a

Ke = 1/2*m*V^2

The Attempt at a Solution

first I calculated the acceleration:

s = (1/2).a.t²

5 = (1/2)a.0.92²

10/0.92² = a = 11.815 ms^-2Then i calculated the net force:

F = m.a

F = 6 * 11.815 = 70.89N

therefore i think friction = 80N - 70.89N = 9.11N Would this be the coefficient of friction?Next i calculated the speed after 0.92s. s = d/t = 5/0.92 = 5.435 ms^-1

Then calculate kinetic energy: = (1/2)m.v² = 3*5.435² = 88.62Jso now would the total energy expended = kinetic energy + Friction force * 5m?

= 88.62 + 9.11 * 5 = 488.65 Nm
can anyone verify this? and let me know if I'm way off please?

This line [red] of your calculation calculates the average speed only. This object is accelerating!

EDIT: And the line before, you calculated the actual friction; not the co-efficient of friction.

 

[oxon88]

ok so because its accelerating, would I use s = (U + V * t) / 2

s = (5.435 *0.92) / 2 = 2.5 ms^-1 ??

 

[oxon88]

is the acceleration not constant though? so surely speed will be constant?

 

[PeterO]

ok so because its accelerating, would I use s = (U + V * t) / 2

s = (5.435 *0.92) / 2 = 2.5 ms^-1 ??

That formula is poorly written, and not appropriate.

This object was accelerating at 11+ m/s/s - if your calculations are correct, and maintaind that acceleration for almost a whole second, so will have reached a speed approaching 11 m/s.

EDIT: that formula should have been written s = (U+V)*t/2

 

[PeterO]

is the acceleration not constant though? so surely speed will be constant?

Speed is constant when the acceleration is zero!

Constant acceleration means the speed continually increases, at a steady rate.

eg: an acceleration of 10m/s² means the speed increases from 0 -10 in one second, then up to 20 m/s [another 10] in the next second, and yet another 10 to 30 m/s after 3 seconds.

That's what acceleration means.

 

[oxon88]

ok i had a re-think.


can i not just use:

Work done = Force x Distance

= 80N x 5m = 400J


Power = Work Done / Time

= 400j / 0.92s = 434.783 Watts

 

[PeterO]

ok i had a re-think.


can i not just use:

Work done = Force x Distance

= 80N x 5m = 400J


Power = Work Done / Time

= 400j / 0.92s = 434.783 Watts

BUT

i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

The original question(s) didn't ask anything about Power ??

 

[oxon88]

ok am i any closer here?...s = (a.t²)/2

5 = a.(0.92²)/2

5 = a.(0.4232)

a = 5/(0.4232) = 11.815 ms^-1

so if the acceleration is 11.815 ms^-1 can i then use v = a.t?V₅ = 11.815*0.92 = 10.87 ms²

v₀ = 11.815*0 = 0 ms²
then do i work out the Ke?

Ke = (1/2)m.v² = (1/2)80x10.87² = 4726.276 J

would this give me the total energy expended?

 

[PeterO]

ok am i any closer here?...


s = (a.t²)/2

5 = a.(0.92²)/2

5 = a.(0.4232)

a = 5/(0.4232) = 11.815 ms^-1

so if the acceleration is 11.815 ms^-1 can i then use v = a.t?


V₅ = 11.815*0.92 = 10.87 ms²

v₀ = 11.815*0 = 0 ms²



then do i work out the Ke?

Ke = (1/2)m.v² = (1/2)80x10.87² = 4726.276 J

would this give me the total energy expended?

Getting closer. when you calculated the Ke, you used 80 for the mass. 80 was the force.

The energy expended will be greater than that, since work was done against friction. That's where part (ii) comes in.

 

[oxon88]

whoops. ok so if i use the mass this time...


Ke = (1/2) 6000 x 10.87²

=354470.7 j

will i need to add this to the energy expended due to friction?


any advice on working out the coefficient of friction?

 

[PeterO]

whoops. ok so if i use the mass this time...


Ke = (1/2) 6000 x 10.87²

=354470.7 j

will i need to add this to the energy expended due to friction?


any advice on working out the coefficient of friction?

Sorry, but the mass is 6 not 6000.

base units in Physics are metres, kg and seconds. I first learned it as the mks system, but better known as SI units [International system of units - but in French where the word for System comes before the word for International]

 

[oxon88]

ok


Ke = (1/2) 6 x 10.87²

= 354.47 j

what else is needed to find the total energy expended?

 

[PeterO]

ok


Ke = (1/2) 6 x 10.87²

= 354.47 j

what else is needed to find the total energy expended?

re-visiting the original post:


i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

(i) I find a little unclear. I could put a case for the answer here being what was requested, and then again the Work done [Your post #7 before you went on pursuing Power] may be what is sought.

Certainly you do a lot of work, and much of that is realized by acceleration to the speed we reached.
The "lost" energy [ie Work - final Ek} will enable you to find the work done against friction - and thus the size of the frictional force. From that you can calculate the co-efficient of friction; μ

 

[oxon88]

re-visiting the original post:


i) calculate the total energy expended in the acceleration

ii) calculate the coefficient of friction between the mass and the surface.

(i) I find a little unclear. I could put a case for the answer here being what was requested, and then again the Work done [Your post #7 before you went on pursuing Power] may be what is sought.

Certainly you do a lot of work, and much of that is realized by acceleration to the speed we reached.
The "lost" energy [ie Work - final Ek} will enable you to find the work done against friction - and thus the size of the frictional force. From that you can calculate the co-efficient of friction; μ

Yes i see your point.

So i could answer the question by saying that the total energy expended is 400 J (Work = F x D)

Then I could go on further and say that the energy expended in the acceleration is 354.47 J, therefore the energy expended because of friction is 45.53 J




can you offer any advice on finding the coefficient of friction?

i have had a go and come up with this:

F=μR_N

R_N=m.g = 6*9.81 = 58.86N

Can i now use the friction I calulated previously (45.53J) in the formula?

45.53 = μ.58.86

μ = 45.53 / 58.86 = 0.774

 

[PeterO]

Yes i see your point.

So i could answer the question by saying that the total energy expended is 400 J (Work = F x D)

Then I could go on further and say that the energy expended in the acceleration is 354.47 J, therefore the energy expended because of friction is 45.53 J




can you offer any advice on finding the coefficient of friction?

i have had a go and come up with this:

F=μR_N

R_N=m.g = 6*9.81 = 58.86N

Can i now use the friction I calulated previously (45.53J) in the formula?

45.53 = μ.58.86

μ = 45.53 / 58.86 = 0.774

Not quite.

Watch you quantities F is a force -measured in Newtons

That 45.53 was 45.53 Joules - a measure of work or energy.

You need to find what distance was covered, so that you can find the size of the force.
OR
Tou could find out the acceleration [I think you already have] and work out what net force is necessary for that - let's say it was 78 N [I just made up that figure]. That would mean that friction must have been 2N, to reduce the effect of the 80N applied force to a mere 78N

Using the correct derived Force, you will get the correct coefficient of friction

If you are worried about change of method on the way through this, you have to become familiar with the fact that these problems can be worked out in a variety of ways - and a kinematics approach may be useful for some values, work/energy for others, impulse/momentum for still others.

Each method can be a good starting point, but usually one method is the quickest and easiest depending what you are trying to calculate.

Note: A few times here you have strayed onto the wrong quantity - and the easiest way to keep track is follow the units.

 

[oxon88]

Ok right so if i say F_Net=m.a

F_Net=6*11.815 = 70.89

then i take that away from the 80N (80N - 70.89 = 9.11N

and then use F = μ.R


9.11 = μ(58.86)

9.11/58.86 = μ

μ = 0.155



so the materials would likely be metal on metal?


Does that look better? Would i not need to take gravity into account when working out F_Net? (somthing like: F_Net= m.a.g)??

 

[PeterO]

Ok right so if i say F_Net=m.a

F_Net=6*11.815 = 70.89

then i take that away from the 80N (80N - 70.89 = 9.11N

and then use F = μ.R


9.11 = μ(58.86)

9.11/58.86 = μ

μ = 0.155



so the materials would likely be metal on metal?


Does that look better? Would i not need to take gravity into account when working out F_Net? (somthing like: F_Net= m.a.g)??

You used gravity when you calculated the Normal reaction force - 58.86N

As for the net force - you don't need gravity, just F[sub[net[/sub] = ma

 

[oxon88]

ok that makes sense. so the answers of μ = 0.155 would be correct?

 

[PeterO]

ok that makes sense. so the answers of μ = 0.155 would be correct?

Looks good. - and is a reasonable answer.