ALL CIE PHYSICS DOUBTS HERE !!!

[Ivo]

What about this question, there seems to be two ways of doing it, getting two different solutions.  I'm not quite sure whether the 1.5 m is required for the calculation, (maybe for GPE?), or do we use the acceleration formula:



The diagram attached shows a mass of 2.5 kg initially at rest on a rough inclined plane.  The mass is now released and acquires a velocity of 4.0ms^-1 at P, the base of the incline.  Find (a) the work done against friction, (b) the (average) friction force.

I got 20J, 5N, but I did I completely different approach (using GPE) and got dissimilar answers: 17.5J, 4.4N.  Which is right - By the way for , can we use this formula if we are working out the friction force (ie. the force against the acceleration), or is this only valid for resultant force?  I'm confused.

Thanks in advance!

[S.M.A.T]

THANKS BRO

[Deadly_king]

What about this question, there seems to be two ways of doing it, getting two different solutions.  I'm not quite sure whether the 1.5 m is required for the calculation, (maybe for GPE?), or do we use the acceleration formula:



The diagram attached shows a mass of 2.5 kg initially at rest on a rough inclined plane.  The mass is now released and acquires a velocity of 4.0ms^-1 at P, the base of the incline.  Find (a) the work done against friction, (b) the (average) friction force.

I got 20J, 5N, but I did I completely different approach (using GPE) and got dissimilar answers: 17.5J, 4.4N.  Which is right - By the way for , can we use this formula if we are working out the friction force (ie. the force against the acceleration), or is this only valid for resultant force?  I'm confused.

Thanks in advance!

Do you mind if I ask how you got 20J and 5N

The only possible answers am getting are 17.5J and 4.375N.

You can use both methods and I assure you, you'll be getting same answers. I guess you must have done some little mistakes somewhere

I'll elaborate on both methods to help you find your errors :

1. Loss in P.E = Gain in K.E + Work done against friction
    mgh = 0.5mv² + Work done against friction
Hence Work done against friction = mgh - 0.5mv² = 2.5(10)(1.5) - 0.5(2.5)(4)² = 17.5J

Now you can find the average frictional force = 17.5/4 = 4.375N

2. First you need to find the angle of inclination(x) using the distances provided -----> sin x = 1.5/4
Now you need to find the acceleration using v² = u² + 2aS
42 = 0 + 2a(4) -----> a = 2m/s²

Using Newton's 2nd law of motion :
Resultant force = ma
mgsin x - Friction = ma
Hence Friction = mgsin x - ma = 2.5(10)(1.5/4) - 2.5(2) = 4.375N

Now, Work done against friction = 4.375 x 4 = 17.5J

Hope it helps

[Ivo]

Do you mind if I ask how you got 20J and 5N

The only possible answers am getting are 17.5J and 4.375N.

You can use both methods and I assure you, you'll be getting same answers. I guess you must have done some little mistakes somewhere

I'll elaborate on both methods to help you find your errors :

1. Loss in P.E = Gain in K.E + Work done against friction
    mgh = 0.5mv² + Work done against friction
Hence Work done against friction = mgh - 0.5mv² = 2.5(10)(1.5) - 0.5(2.5)(4)² = 17.5J

Now you can find the average frictional force = 17.5/4 = 4.375N

2. First you need to find the angle of inclination(x) using the distances provided -----> sin x = 1.5/4
Now you need to find the acceleration using v² = u² + 2aS
42 = 0 + 2a(4) -----> a = 2m/s²

Using Newton's 2nd law of motion :
Resultant force = ma
mgsin x - Friction = ma
Hence Friction = mgsin x - ma = 2.5(10)(1.5/4) - 2.5(2) = 4.375N

Now, Work done against friction = 4.375 x 4 = 17.5J

Hope it helps

Thanks for your help.  I understand the 1st method, but I'm not sure about the 2nd.  How did you get mgsin x?  I assume you're taking angle x is the point at P on the diagram - so for the angle:



But then how do we work out the acclerating force (how did you get your value)?  I'm assuming that the weight of 2.5kg mass is acting vertically downwards, so it is 25N.  So I take the acclerating force to be:

, where F is the accelerating force to be found?

Many thanks once again in advance!

[Freaked12]

you know initial and final velocities
 and distance travelled
 find acceleration


2 ms-2
 calculate the force of weight acting downwards
thats coming 3/8

this force minus friction gives you ma
 you know m and a
 equate and find friction
 and then multiply it by distance, 4m
 and you'll get work done

[Ivo]

you know initial and final velocities
 and distance travelled
 find acceleration


2 ms-2
 calculate the force of weight acting downwards
thats coming 3/8

this force minus friction gives you ma
 you know m and a
 equate and find friction
 and then multiply it by distance, 4m
 and you'll get work done

How did you get 3/8?  Surely force acting downards is 25N, no?

[Freaked12]

sorry
 thats 3/8 into 25
 thats it
 FsinQ
 sinQ = 3/8

[Ivo]

sorry
 thats 3/8 into 25
 thats it
 FsinQ
 sinQ = 3/8

I agree that , and that (Accelerating) , which means the .  But the answer gives 66.7N, which is not right.  Surely from Deadly_king's answer, then the accelerating force must be 9.4N instead, so that the friction force = 4.4N, as said by Deadly_king.

Thanks in advance!

[Deadly_king]

Thanks for your help.  I understand the 1st method, but I'm not sure about the 2nd.  How did you get mgsin x?  I assume you're taking angle x is the point at P on the diagram - so for the angle:



But then how do we work out the acclerating force (how did you get your value)?  I'm assuming that the weight of 2.5kg mass is acting vertically downwards, so it is 25N.  So I take the acclerating force to be:

, where F is the accelerating force to be found?

Many thanks once again in advance!

Indeed the weight acts downwards and is equal to 25N. But for this question you need not the weight but the component of the weight along the plane,i.e the inclined plane.

If you draw a straight line vertically downwards from the centre of the mass, the latter will form a right angled triangle with the horizontal line. From this you find the angle of inclination which is sin x = 1.5/4.
Now you can find the component of the weight downwards along the plane which will be 25sinx or 25(1.5/4)

NOTE : Angle x which is equal to the angle of inclination is also the angle between the vertical and the perpendicular line of the plane from the centre of mass.

Try to draw the triangle and the normal, you'll see that the angle I mean is equal to x and also how I got 25sin x as component of the weight downwards but along the plane.

Hope you understand what am trying to say  

[Ivo]

Indeed the weight acts downwards and is equal to 25N. But for this question you need not the weight but the component of the weight along the plane,i.e the inclined plane.

If you draw a straight line vertically downwards from the centre of the mass, the latter will form a right angled triangle with the horizontal line. From this you find the angle of inclination which is sin x = 1.5/4.
Now you can find the component of the weight downwards along the plane which will be 25sinx or 25(1.5/4)

NOTE : Angle x which is equal to the angle of inclination is also the angle between the vertical and the perpendicular line of the plane from the centre of mass.

Try to draw the triangle and the normal, you'll see that the angle I mean is equal to x and also how I got 25sin x as component of the weight downwards but along the plane.

Hope you understand what am trying to say  

Thanks, for your help.  Great explanation, now I get it!  + rep for you!  (By the way, which method would you advise?)

[Deadly_king]

Thanks, for your help.  Great explanation, now I get it!  + rep for you!  (By the way, which method would you advise?)

You're welcome

Hmm.....that will depend on you. Use the method you find easier.

But the question ask for the work done first, so i believe the first method would be shorter

[Ivo]

OK, I think this is slightly harder than the previous ones that I am struggling with.  I know that the tension is the same all round the system, but I don't know which directions each one are (could someone draw the arrows to show them, thanks!)

My (wrong) answers:

a) For equlibrium to be achieved, forces are balanced, zero resultant.

b) i) 141N

    ii) 131N

    iii) No idea!  Difference is not 0N, I'm stuck!

Many thanks in advance!

[S.M.A.T]

I got the same answer like U in b) i) ii) ....what is the correct answer

b iii)because the frictional force in the bed is equal to the horizontal force exerted by the rope so resultant force is zero

[ashish]

OK, I think this is slightly harder than the previous ones that I am struggling with.  I know that the tension is the same all round the system, but I don't know which directions each one are (could someone draw the arrows to show them, thanks!)

My (wrong) answers:

a) For equlibrium to be achieved, forces are balanced, zero resultant.

b) i) 141N

    ii) 131N

    iii) No idea!  Difference is not 0N, I'm stuck!

Many thanks in advance!

 your answers are not wrong

i got the same

look at the picture

there is a constant tension(10* in the string throughout the system i took g as ten

when the string has inclined it will cause an additional force

additional horizontal force will be 80cos40 =61.3
total force will be                                =61.3+80
                                                        =141 N

additional vertical force will be 80sin40 = 51.42
total force will be                             = 51.42+80
                                                     =131.42 N



  

[Deadly_king]

I got the same answer like U in b) i) ii) ....what is the correct answer

b iii)because the frictional force in the bed is equal to the horizontal force exerted by the rope so resultant force is zero

Do you mind to explain how you got your answers

b)(iii) It's not necessary that resultant is zero. The frictional force between the bed and the body of the skier may be greater or equal to the horizontal force.

@ ashish : Weight cannot be the answer since the latter acts vertically downwards ans will have no horizontal component