Since the coefficient of friction is \(\frac { 1 }{ \sqrt { 3 } } \)
tanθ = \(\frac { 1 }{ \sqrt { 3 } } \) ⇒ θ=30⁰.

Angle of inclination θ = 45⁰
∴ Coefficient of friction μ= tan θ = tan45⁰=1

To find the force, we need to find the acceleration experienced by the particle.
Th e acceleration is given by a=\(\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } \) or a=\(\frac { dv }{ dt } \)
Here v = velocity of the particle in y direction
v = \(\frac { dy }{ dt } \) = u - gt
The momentum of the particle = mv = m(u - gt)
a = \(\frac { dv }{ dt } \) = -g
The force acting on the object is given by F = ma = -mg
The negative sign implies that the force is acting on the negative y direction. This is exactly the force that acts on the object in projectile motion.

The velocity of the particle is \(\vec { v } ={ v }_{ x }\hat { i } +{ v }_{ y }\hat { j } +{ v }_{ z }\hat { k } \\ \). As shown in the figure, the particle is moving in the xy plane, there is no motion in the z direction. So velocity in the z direction is zero (v_z = 0). The velocity of the particle has x component (v_x) and y component (v_y). From figure, as time increases from t = 0 sec to t =3 sec, the length of the vector in y direction is changing (increasing). It means y component of velocity (v_y) is increasing with respect to time. According to Newton's second law, if velocity changes with respect to time then there must be acceleration. In this case, the particle has acceleration in the y direction since the y component of velocity changes. So the particle experiences force in the y direction. The length of the vector in x direction does not change. It means that the particle has constant velocity in the x direction. So no force or zero net force acts in the x direction.

From Newton's second law (in magnitude form), F = ma.
For the object of mass 2.5 kg, the acceleration is a =\(\frac { F }{ m } =\frac { 5 }{ 2.5 } \) = 2 ms^-2
For the object of mass 100 kg, the acceleration is a = \(\frac { F }{ m } =\frac { 5 }{ 100 } \)= 0.05 ms^-2