If a particle of mass m accelerates in the x direction, the force in x direction is given by

(according to Newton’s 2nd law of motion).

**Ʃ** F _{x} = m a _{x} [ scalar equation ]

**Ʃ **F _{y} =0, Ʃ F _{z} =0

In general case if the particle of m, has components a_{x}, a_{y} and a _{z} in three directions, we have

**Ʃ** F _{x} = m a _{x }

**Ʃ** F _{y} = m a _{y}

**Ʃ** F _{z} = m a _{z}

**UNIT OF FORCE :** S.I. unit of force is Newton (N).

l N = l kg x l m/s2 = 1 kg m/s^{2}.

**SOLVED EXAMPLES**

**Example 5.1.** A force acts on a body of mass 15 kg and produces an acceleration of 6 m/s2 in the direction of force. Determine_ tlze magnitude of the force.

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**Solution.** Given m =· 15 kg, a = 6 m / s2

Force F= ma

=15 x 6kg m/s^{2}.

= 90 N. Ans

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**Example 5.2.** A force of 1000 N acts on a body having mass of 8 kg for 10 seconds. If the initial velocity of the body is 10 m/s, determine (i) acceleration produced in the direction of force, and (ii) distance moved by the body in 10 seconds.

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**Solution.** F= ma

1000 = 8 x a

a = 1000 / 8 = 125 m / s^{2 }

v_{0} =10 m /s ,t =10 + 125 x10 = 1260 m / s^{2}

v = v_{0} +at = 10 + 125 x 10 = 1260 m/s

v^{2} = v_{0} +2 x a x s

(1260)^{2} =(10)2 +2 x 125 x S

S = (1260)^{2} – (10)^{2} / 2 x 125 = 6350 m.

**Example 5.3.** The weight of a body on earth is 981 N. If the acceleration due to gravity on earth is 9.81 m/s^{2}, what will be weight of the body on (i) moon (g moon = 1.7 m/s2), (ii) Sun (g sun = 275 m/s^{2}).

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**Solution. **W = mg

981 = m x 9.81

m = 981 / 9.81 =100 kg

(i) The mass remains constant.

W at moon= m x g moon

= 100 x 1.7 = 170 N. Ans.

(ii) Weight at sun

W = m x g sun

=100 x 275 27500 N. Ans.

**Example 5.4.** A force of 20 N acts on a body having mass of 30 kg for 90 seconds. If the initial velocity of the body is 2 m/s, determine the final velocity of the body when (i) the force acts in the direction of motion ; (ii) the force acts in the opposite direction of motion.

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**Solution :** 20 = 30 x a

a = 2/3 m/s2

v = v_{0} + at = 2 +2 /3 x 90 = 62 m/s .Ans.

(ii) force = -20

a = -20 /30 = -2 / 3

v = v_{0 } +at = 2 / 3 x 90 = – 58 m / s . (negative sign shown

the body will move in the direction of the force which is opposite of case (i))

**Example 5.5.** A body of mass 30 kg falls from rest on the ground from a height of 20 m. If the resistance of the ground against penetration is 5000 N, find tl1e depth of penetration.

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**Solution. **v_{0} = v_{0}^{2} + 2g x 20

= 0 + 2 x 9.81 x 20

= 19.81 m/s.

Weight of the body, W = mg = 30 x 9.81

= 294.3 N.

Net upward force = 5000 – 294.3

= 4705.7 N ; 4705.7 / 30 = a (retardation)

v^{2} = v_{0}^{2} – 2aS

0 = (19.81)2 – 2 x 4705.7 /30 x S

S = 1.251 m. depth of penetration .Ans.