Given : Mass, m = 3 kg;

Initial Velocity, u = 2m/s;

Final velocity, v = 3.5 m/s

Time taken, t = 10 seconds

v = u+at

3.5 = 2 +a x 10

F = ma = 3 x 0.15 = 0.45 N

Since the applied force increase the speed of the body, it acts along the direction of motion.

The product of velocity and mass is known as momentum which is a vector quantity.

For SI to CGS conversion,

1 Kg = 10³ g

1 m = 10² cm

=> In CGS System,

1 N (Kg.m/s²) = 105 dynes (g.cm/s²)

=> 1 g.wt = 9.8 m/s² x 10² cm / 1 s² x 10^-3 g x 10^-2 cm / 1 s x 10⁵ dynes

= 980 dynes (g.cm/s²)

Planck's Constant (h)  = 6.626176 x 10-34  m2 kg/s
So, Unit of planck constant=  m² kg/s

Dimensions =M L² T ⁻¹   ________ (1)

Angular momentum l = mvr

Where, m-mass

v-velocity

r-radius

Dimensions of angular momentum  =  M L T ⁻¹ L   = M  L²  T ⁻¹  _______________ (2)
From (1) and (2). 

Planck's constant and angular momentum have the same dimensions.

Net external force F = dp/dt, so using dimensional analysis, unit of momentum should be the same as unit of force times unit of time which is Newton second.

1N = kg m /sec
1 dyne = gram cm/sec
Thus 1N in CGS system = 1000 gram 100 cm/sec
That is 1N = 100000 dyne

Force vector follows the principle of superposition which says all the force vectors can be vectorially added if applied on one point to get the net force vector. Hence we get
F = F₁ + F₂ + F₃ 
= (-2 + 3) i + (1 - 2) j
 F = i - j = ma
Thus we get a = (i - j) /m
= (i - j) / 2

Newton's 2nd Law of motion states that rate of change of momentum is directly proportional to external force applied. Hence, it relates rate of change of momentum to external force applied.

F = ma so a= 250/30

a = 8.34

now the result is 4mt/secsqure so reduction is 4.34

so friction force will b 4.34 ×30

 F is 130.2 n

Time rate of change of angular momentum represents torque. Hence force is known as rate change of momentum. The rate of change of momentum is called force by Newton's second law of motion.