Mathematics: Post your doubts here!

crazytaylorfanXD

crazytaylorfanXD

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TheJDOG said:
Hi,
5)
i) they want the KE gain of the whole system, we know that KE=1/2mv^2
We know block A mass= 5 kg and B= 16 kg
So KE of whole system = (1/2)5v^2 + (1/2)16v^2 = 10.5v^2 J

ii) a) we need the loss of PE of the system. We know that PE= mgh and we know they moved a distance of x
We also know that the loss of PE of the system= Loss of b - gain of A
Well, logically B lost PE it moved downwards, so PE Loss of b= mgh= 16(10)(x) J
Also, A gained PE it moved up the slope, use basic trigonometry to find the distance it moved vertically up in terms of x.
So PE gain of A= mgh= 5(10)(xsin30) J
Loss of PE of the system= 16(10)(x) - 5(10)(xsin30)= 160x - 25x= 135x J

b) here we need the work done against friction. We know that f=uR and work = FxD
given u= 1/root3
We have to resolve, take y axis perpendicular to slope. We have R acting upwards along y axis and we have the component of weight of A acting downwards along y axis. Now R= Wcos30=50cos30= 25root3.
So friction = uR= (1/root3)(25root3)= 25 N
So work done against friction = 25x J since it moved a distance of x

iii) we know that Gain in KE= Loss in PE - work done to oppose friction
So 10.5v^2= 135x - 25x
10.5v^2 = 110x
Multiply both sides by 2
21v^2= 220x
Click to expand...


thank you so much :)
 
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ashcull14

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manya said:
can someone help me with this question
while calculating work done by driving force y dont we include (weight x distance) as workdone by frictional force
http://www.themusicofmath.com/wp-content/uploads/2012/09/Oct.-Nov.-2008-Paper-4-Mechanics-1.pdf
Click to expand...

rules for inclined plane
gravitational potential energy= component of weight (mgsin0) *distance
Kinetic energy 1/2mv^2-u^2 OR resultant force*distance
WD BY FRICTION = FRICTION (RESISTIVE FORCE )*DISTANCE (ALWAYS) because we r calculating the work done due to friction NOT weight....as WD=FORCE*distance ....u can judge it by the term WD BY FRICTION ...its not WD due to WEIGHT ...hope u got it
so the formula 0=
wd by driving force= kinetic energy (1/2mv^2-u^2)+wd by friction (friction *distance)+ GPE (mgsin0*distance)
 
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ashcull14

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manya said:
can someone help me with this question
while calculating work done by driving force y dont we include (weight x distance) as workdone by frictional force
http://www.themusicofmath.com/wp-content/uploads/2012/09/Oct.-Nov.-2008-Paper-4-Mechanics-1.pdf
Click to expand...
if ure tlking about que 4
here no driving force is req to be calculated neither is any friction stated
the object is being lifted (height is increasing) means GPE increases
the speed increases fron 0m/s (rest) to 0.5m/s so kinetic energy increases
so WD = Gain in KE + Gain in GPE
m=160kg
v=0.5m/s
u=0m/s
s=?
h = (v+u/2)*t
(0+0.5)/2 * 7 = 1.75m
so GPE = mgh = 160*10*1.75 =2800J
KE= 1/2mv^2-u^2 = 0.5*160* (0.5^2-0^2)= 20J
so total WD= 2800+20 = 2820J
 
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ashcull14

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m1 nov 2008.PNG
just have a bit of confusion ...how was c found when acceleration was integrated? nd why did we use 2.5s as time to find the total distance??
 
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