Qualification > Sciences
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 ;D ;D
Deadly_king:
--- Quote from: Ivo on October 15, 2010, 10:45:09 pm ---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!
--- End quote ---
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:
--- Quote from: Deadly_king on October 16, 2010, 05:00:30 am ---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 :)
--- End quote ---
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
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