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8 years ago

```							total change in height = dh =H-R =50cm=0.5m
from work energy theorem,
change in kinetic energy = total work
kinetic energy = mgdh
mv2(1+k2/r2)/2 = mgh
mv2= 2mgh/(1+k2/r2)    .....................1
k is radius og gyration & for sherical ball it is Rsqrt(2/5) ...
mv2 = 10mg/7           .................2
now when theta =0 then
N  = mv2/R
so N = 10mg/7R=10Newton
since the body is not falling off so net friction force should be equal to weight =mg
f = 0.7newton
```
8 years ago
```							thanks for the reply.... normal reaction is correct but frictional force is wrong by ur answer..... the right answer is 0.2 N

sorry ,i forgot to tell that it is small sphere that is is rolling down the inclined plane ....

i did it this way (but failed)
frictional force = f , alpha=angular acceletion
at  point A
f - mg = ma..... (1)
f . r = I (alpha)....(2)
putting f = I (alpha)/r in (1)
I(alpha)/r=m(g+a)
for sphere I=2mr(square)/5 and alpha=a/r
so
2mr(sqr)/5 * a/r * 1/r=m(g+a)
2/5 a= g+a
a=-5/3g
putting this in (1)
i got f=0.46

plz remove my confusion
thanks
```
8 years ago
```							 eq 1 should be
mg - f = ma ..........1
(this is because ,direction of net force on center of mass is downward ...)
f = Ia/r2
mg = 7f/2
f = 2mg/7 =0.2N ans
please approve if u like my ans
```
8 years ago
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