M2 Work Energy and Power Topic

Original post by potato343

Hey everyone, i'm having some trouble with a question which is:

A Crate of Mass 40kg is pulled by a rope at a constant 1.5 m/s down a slope inclined at 15 degrees to the horizontal. Contact is rough and the co-efficient of Friction is 0.7. Find:

a) the frictional force
b) the tension in the rope
c) the work done by the rope per second
d) the work done by gravity while the crate moves down the slope for 6 seconds.

Note: they tell me to take gravity as 10 and to round my answer to 2 significant figures.

The answers they give are:

a)270N
b)170N
c)250N
d)930N

I am seriously struggling with this as they have not given me the distance of the slope. Also the answers they give so that I can check if I was correct seem impossible to get to.

Thanks very much

a) To find frictional force, you use F = mR (m being the coefficient of friction)
so it is 0.7 x 40gcos15 = 270.5
rounding it off to two sig.fig gives 270

b) since it's moving at constant velocity, t + 40gsin15 = frictional force
t + 40gsin15 = 270
t = 166 and again, rounding off gives 170

d) well you're told that the speed is 1.5m/s, so in 6 seconds it travels a distance of 9m

gravitational potential energy = mgh so 40 x 10 x 9sin15
(there is a sin15 involved because you need the vertical distance it travelled, not the distance down the slope)
= 931 which is rounded to 930

Just realised i skipped part c, I'll have to check that one as i'm not too sure how to do it at the moment :smile:

(edited 12 years ago)

Reply 2

Original post by potato343

Hey everyone, i'm having some trouble with a question which is:

A Crate of Mass 40kg is pulled by a rope at a constant 1.5 m/s down a slope inclined at 15 degrees to the horizontal. Contact is rough and the co-efficient of Friction is 0.7. Find:

a) the frictional force
b) the tension in the rope
c) the work done by the rope per second
d) the work done by gravity while the crate moves down the slope for 6 seconds.

Note: they tell me to take gravity as 10 and to round my answer to 2 significant figures.

The answers they give are:

a)270N
b)170N
c)250N
d)930N

I am seriously struggling with this as they have not given me the distance of the slope. Also the answers they give so that I can check if I was correct seem impossible to get to.

Thanks very much

First things first draw out a nice clear diagram, label everything and resolve the weight in components.

A) Resolve perpendicular to the plane and use F=uR (u = coefficient of friction)

B) Resolve parallel to the plane and use the answer to A)

C) Its not looking for the Work done, it's looking for the work done per second, I.e) Power.

D) Remember: Change in P.E = mg x (vertical height change) and S = Ut..

Reply 3

X

It's best not to do the work for them but guide them in the right direction.

Reply 4

Original post by gozatron

It's best not to do the work for them but guide them in the right direction.

Whoops, I've got the same exam coming up so i just wanted to make sure i was getting the workings all correct :redface:

Although how would you have worked out part c? I realise it's power they're asking for but i dont seem to be getting the correct answer

Reply 5

Whoops, I've got the same exam coming up so i just wanted to make sure i was getting the workings all correct :redface:

Although how would you have worked out part c? I realise it's power they're asking for but i dont seem to be getting the correct answer

No worries, I know how you feel I'm taking M2 aswell. :p:

Do you know the formula for power?

Wd = F x S
P = (F x S)/t = F x V

Can you take it from there?

Reply 6

Original post by gozatron

No worries, I know how you feel I'm taking M2 aswell. :p:

Do you know the formula for power?

Wd = F x S
P = (F x S)/t = F x V

Can you take it from there?

I did exactly that, although I used the tension as my F?
And ah, edexcel board?

Reply 7

I did exactly that, although I used the tension as my F?
And ah, edexcel board?

Edexcel

Yep you use tension, did it not work? F=170N, V=1.5m/s?

Reply 8

Original post by gozatron

Edexcel

Yep you use tension, did it not work? F=170N, V=1.5m/s?

Well apparantly the exam is either going to be ridiculously easy or ridiculously hard, according to a couple of examiners..I know which one i'm hoping for :redface:

And yeah, 255 right? I may be a complete idiot here, but does that not round to 260 and not 250?

Reply 9

potato343

OP

a) To find frictional force, you use F = mR (m being the coefficient of friction)
so it is 0.7 x 40gcos15 = 270.5
rounding it off to two sig.fig gives 270

b) since it's moving at constant velocity, t + 40gsin15 = frictional force
t + 40gsin15 = 270
t = 166 and again, rounding off gives 170

d) well you're told that the speed is 1.5m/s, so in 6 seconds it travels a distance of 9m

gravitational potential energy = mgh so 40 x 10 x 9sin15
(there is a sin15 involved because you need the vertical distance it travelled, not the distance down the slope)
= 931 which is rounded to 930

Just realised i skipped part c, I'll have to check that one as i'm not too sure how to do it at the moment :smile:

Why when working out the reaction, is it 40gcos15, surely its 40g/cos15 because cos=A/H ? this confuses me the most

Reply 10

Original post by potato343

Why when working out the reaction, is it 40gcos15, surely its 40g/cos15 because cos=A/H ? this confuses me the most

Because when you draw the diagram, the component of the weight that is acting in the opposite direction to the reaction is 40gcos15 :smile:

Reply 11

F1Addict

12

Well apparantly the exam is either going to be ridiculously easy or ridiculously hard, according to a couple of examiners..I know which one i'm hoping for :redface:

And yeah, 255 right? I may be a complete idiot here, but does that not round to 260 and not 250?

If you use 170, you get 255, which rounds to 260. I think they used 166 instead of 170, which gives 249, which rounds to 250.

Reply 12

Original post by F1Addict

you get 255, which rounds to 260. I think they use 166 instead of 170, which gives 249, which rounds to 250.

Ahhh, thanks for clearing that up :smile:

Reply 13

Well apparantly the exam is either going to be ridiculously easy or ridiculously hard, according to a couple of examiners..I know which one i'm hoping for :redface:

And yeah, 255 right? I may be a complete idiot here, but does that not round to 260 and not 250?

Oh god, I hope for a good paper! I haven't done many past papers so I'm kinda worried in that respect :p:

Yeh your right thats what I get for using rounded numbers :facepalm:

It does work though if you use more sig dogs or g as 9.8 :p:

Reply 14

Original post by gozatron

Oh god, I hope for a good paper! I haven't done many past papers so I'm kinda worried in that respect :p:

Yeh your right thats what I get for using rounded numbers :facepalm:

It does work though if you use more sig dogs or g as 9.8 :p:

The past papers have all been fairly okay, aside from some dodgy questions which come up from the last chapter :frown:

And I did the whole question initially using g = 9.8 (because i'm so used to it) so when i reread the question i had to do it all over again using g = 10, it's what i get for never reading questions properly i suppose :colondollar:

Reply 15

F1Addict

12

Original post by gozatron

Oh god, I hope for a good paper! I haven't done many past papers so I'm kinda worried in that respect :p:

Yeh your right thats what I get for using rounded numbers :facepalm:

It does work though if you use more sig dogs or g as 9.8 :p:

I reckon it'll be a hard paper. In a way its better if its hard. If you work at M2 and can answer most questions then you'll probably end up getting more UMS that on an easy paper where one silly mistake can cost you a grade or two. :tongue:

I'm not worried. I could fail M2 and still get an A/A*. It all depends on C4, which can be a pain.. =/

Reply 16

potato343

OP

Because when you draw the diagram, the component of the weight that is acting in the opposite direction to the reaction is 40gcos15 :smile:

But i thought that weight is straight down, and the reaction is perpendicular to the inclined plane, not directly opposite each other
:s-smilie:

Reply 17

Original post by potato343

But i thought that weight is straight down, and the reaction is perpendicular to the inclined plane, not directly opposite each other
:s-smilie:

Exactly, so the component of weight in the opposite direction of the reaction is mgcosx
I'm pretty sure this was covered in M1?

Reply 18

potato343

OP

Exactly, so the component of weight in the opposite direction of the reaction is mgcosx
I'm pretty sure this was covered in M1?

Yeh it was, i'm just really getting myself into a cafuffle, bout trigonometry and the like. It doesn;t bode well :/

Reply 19