Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
Ivo:
--- Quote from: Requiem on October 16, 2010, 02:40:53 pm ---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
--- End quote ---
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:
--- Quote from: Requiem on October 16, 2010, 03:05:47 pm ---sorry
thats 3/8 into 25
thats it
FsinQ
sinQ = 3/8
--- End quote ---
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:
--- Quote from: Ivo on October 16, 2010, 01:49:58 pm ---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!
--- End quote ---
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:
--- Quote from: Deadly_king on October 16, 2010, 05:14:45 pm ---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 ;)
--- End quote ---
Thanks, for your help. Great explanation, now I get it! + rep for you! (By the way, which method would you advise?)
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