Class 11 Physics Laws of Motion Newtons Second Law

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

Fx = dpx/ dt = max

Fy = dpy/ dt = may

Fz = dpz/dt = maz

• 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

Solution.

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

Or,  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.

Solution.

m= 5kg

Fx = 15N, Fy = 25N

F = Fxi + Fyj

= √ 225 + 625 = √850

F = ma

Or, a = F/ m = √850/5 = 5 √34/5 = √34 = 5.83m/s2

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