Both the methods are equally good and applicable in solving spring related problems.

 

May be u are committing some fundamental mistake while solving such problems.

 

You may post the solution by different methods for solving same problem in which you are encountering the different results, so that we can suggest you the lacunae, if any.

 

Moreover chirag has rightly stated that both metnods converges to the same result and can be derived from the other.

Mass of earth is =Me =  5.98 x 10²⁴ kg

 

Due to this mass there is a gravitational force of attraction between earth and objects on the earth.

 

So, acceleration due to gravity on earth's surface is defined as g = GMe / R²

 

Where, R is radius of earth ~ 6400km = 6.4 x 10⁶ m

 

or, g = GMe / R² = 9.8 m/s²

 

Now keeping the same fact in mind if we try to asscociate weight of earth it meas

 

Weight of earth = mass of earth x g'

 

where g' is accleleration on earth due to bodies present in its vicinity as well due to other celestial bodies like sun etc.

 

But the bodies on earth are having masses too small that contributes negligibly towards g'

 

Similarly if we try to find out the contribution from SUN towards g' then

 

we see that g' = G (mass of sun)/ (distance between sun and earth)²

 

= G Ms / R_es²

 

Here  R_es = distance b/w earth and sun = 1.52 x 10¹¹ (Aphelion)

 

also  R_es = 1.47 x 10¹¹ ( At the time of Perihelion)

 

Thus if we take R_es = distance b/w earth and sun = 1.52 x 10¹¹

 

G = 6.64 x 10^-11

 

Ms = 1.98 x 10³⁰ kg

 

then we find that g' ~ 10^-9

 

Which is far less and negligible.

 

Weight of earth due to sun = Me x g' ~ 12 x 10¹⁵ N

 

thus in this way the weight of earth can be associated and corelated due to the prescence of celestial or other heavy bodies in its vicinity.

 

Though there is no specific and standard idea to estimate weight of the earth due to ambiguity involved in the preoblem as evident from above analytical but crude example.

Your third question is

 

Q).the frictional force does not increase by increasing  the surface area of the tyre of the vehicle but in reality why such tyres experience more friction.

 

Ans) If we talk about sliding friction then,

 

since frictional force is given by

 

f =  N

 

where,  is coeff. of friction and

 

N is normal reaction force by the road on the vehicle.

 

So here there is no term like surfac area.

 

thus the frictional force is independent of surface area of the vehicle.

 

In a simple system in equilibrium, such as a mass sitting on a surface, the

 

normal force is equal to mass multiplied by gravitational acceleration (F_N = mg).

 

The statement of this expression is that friction is directly proportional to the

 

intrinsic friction caused by the materials and contaminants; and that friction is

 

also directly proportional to the force. In practice, this is an approximation but in

 

many situations other factors, e.g., the area of contact, play a role.

 

 

Answer to the first question is

 

The third differential of dispalcement 'x' is

 

d³x/dt³ = da/dt (where a = acceleration)

 

thus, acceleration is changing w.r.t time means it is tha case of

 

VARIABLE FORCE.

 

as F = ma and since a is changing, means force on the system is changing w.r.t time.

 

Or it can be termed as JERK.

Q)  The electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second. find the maximum KE of the photoelectrons when this light falls on a metal surface  whose work function is 2eV.

 

Solution)

 

 

 

 

 

thus,   eV₀ = hf -

 

 

 

 

 

 

 

 

 

For PHYSICS:             http://www.fiitjee.com/down/sol/physice06.pdf

 

For CHEMISTRY:        http://www.fiitjee.com/down/sol/chemistry06.pdf

 

For MATHEMATICS:    http://www.fiitjee.com/down/sol/maths06.pdf

 

Wishing you a very good luck.

Anyway if chages are +q and -q separated by some distance r.

 

Then if say charge +q/2 is palced in between as shown below

 

 

+q.....................+q/2......................-q

 

Then force on q/2 due to q is

 

F1 = 2kq²/r² from q to q/2

 

similarly force on q due to -q/2 is

 

F2 = -2kq²/r² but in the direction opposite to F1

 

Thus total force on teh charge = F1 -F2 = 0

 

Thus resultant force = 0

Yes troy and arjun have rightly argued that

 

sin₁₊sin₂+sin₃=3 is possible only when ₁=₂=₃=/2

 

thus cos₁ = cos/2 = 0

 

 

But w.r.t earth accleration of ball =  [(horizonatal)² + (vertical)²]

 

So acceleartion w.r.t earth =  a² + g²

 

and angle at which the ball is falling is given by

 

 tan  = g/a   where  is the angle w.r.t vertical drawn from the point from where ball is dropped.