i think the above calculations by me and vasanth r wrong

coz

by conservation of energy,we hv

 

mgx = 1/2 mv²

 

x = 1/2*4/10 = 0.2 m

 

I DUNNO WHICH METHOD HAS GOT THE RUDDY "FLAW"

SOMEONE FIND OUT THE MISTAKE

 

Either there has to be infinite numbers of ordered pairs or  none
coz 7/15 is irrational equal to 0.466666666666666666666666666666666

 

 

 

Hi..

for the body to just slide

 

Fcos 60 = uN

solving we get

 

F ( 1/2 - 0.1{sqrt3} )  = sqrt3  [ assuming the mass is sqrt3 ]

 

F = 5.29 N

ANSWER IS (D)

Hope this is clear..

I.m trying out the first qu

 

firstly at equlibrium,

                             mg = kx

                            so, k=500 N/m

Then,when the impulse is given the equation is,

 

kx - mg = -ma

            = -m dv/dx dx/dt

 

(kx - mg) dx = - mv dv

 

integrating,

 

 

_[o]^[x] kx dx - mg _[o]^[x]  dx = - m _[2]^[0] v. dv

applying limits we get

 

kx² - 2mgx + mv²  = 0

 

solving this I get

 

x = 0.1 m

 

If thats rite plz a salute and if its wrong please point out the mistake coz i think the method is rite

 

I think the answer is wrong

firstly the mass will hv some velocity at the upper end which is equal to

sqroot(gl)   [which is the critical velocity to complete a vertical circle...]

 

energy eqn now changes to

 

1/2 m v² = 2lmg + 1/2 mgl

 

From ths we get v = sqrt (5gl)

 

If i'm wrong dont forget to point out the mistake and if rite a salute....

Take the effect of bulging of earth and its rotation in account.Consider the following statements

 

(A) There are points outside the earth where the value of g is equal to its value at the equator.

(B) There are points outside the earth where the value of g is equal to its value at the poles.

 

Ans:(A) is right but (B) is wrong

 

I want a clean,clear mathematical explanation.....

I think its quite simple

first differentiate the expression dv/dt=k

so, a=2m/s²

now use s= ut + 1/2 at²

                  s= 0 + 1/2 * 2 * (3)²

              

              s= 9m

plz rate me