Newton’s Second Law of Motion

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  • Momentum
  • Newton's second law of motion
  • Apparent weight of a body in a lift


Newton’s Second Law

  • The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

  • Alternatively, the relationship between an object's mass m, its acceleration a, and the applied force F is F = ma; the direction of the force vector is the same as the direction of acceleration vector.
    `F ∝ (dp)/dt` [Greater the change in momentum, greater is force]
    `F = k (dp)/dt`
    `F = (dp)/ dt`
    `F = d/ dt (mv)`
    Let, m: mass of the body be constant
    `F = m (dv)/dt`
    `F = ma`

  • Newton’s Second law is consistent with the First law
    F = ma
    If F = 0, then a = 0
    According to First law, if a = 0, Then F = 0
    Thus, both the laws are in sync.

  • Vector form of Newton’s Second law
    `bar F=barF_x hat i + bar F_y hat j+ bar F_z hat k`

    `F_x = "dp"_x/ dt = ma_x`

    `F_y = "dp"_y/ dt = ma_y`

    `F_z = "dp"_z/dt = ma_z`

  • Newton’s Second Law was defined for point objects. For larger bodies,

    a: acceleration of the centre of mass of the system.

    F: total external force on the system

Problem 1: A car of mass 2 × 103 kg travelling at 36 km/hr on a horizontal road is brought to rest in a distance of 50m by the action of brakes and frictional forces. Calculate: (a) average stopping force (b) Time taken to stop the car

m = 2 × 103
u = 36 km/hr = 10m/s
s = 50m
v = 0

To find: a, F, t
Third equation of motion:
v2 = u2 + 2as
0 = 100 + 2a × 50
a = -1 m/s2

Therefore, F = ma = (2 × 103) × 1 = 2 × 103 N

V = u + at
0 = 10 – t
t = 10 sec

Problem 2: The only force acting on a 5kg object has components Fx = 15N and Fy = 25N. Find the acceleration of the object.
`m= 5kg`
`F_x = 15N, F_y = 25N`
`F = F_x hat i + F_y hat j`
`| bar F|=sqrt(F_x^2 + F_y^2)= sqrt( 225 + 625) = sqrt 850`

`F = ma`
Or, a = `F/ m` = `sqrt 850/5= 5 (sqrt34)/5= sqrt34= 5.83 m/s^2`


  • Impulse is defined as a force multiplied by time it acts over.

  • For example: Tennis racket strikes a ball, an impulse is applied to the ball. The racket puts a force on the ball for a short time period.

    `F ∆t = ∆p`

    `F = (∆p)/( ∆t)` = Rate of Change of momentum

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