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

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 * 10^{3} 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 * 10^{3}

u = 36 km/hr = 10m/s

s = 50m

v = 0

To find: a, F, t

Third equation of motion: v^{2} = u^{2} + 2as

0 = 100 + 2a * 50

Or, **a = -1 m/s ^{2}**

Therefore, F = ma = (2 * 10^{3 }) * 1 = **2 * 10 ^{3} N**

** **V = u + at

0 = 10 – t

** t = 10 sec**

**Problem 2: **The only force acting on a 5kg object has components F_{x} = 15N and F_{y} = 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/s ^{2}**

.