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    Laws of Motion
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    A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–1 to 3.5 m s–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
    Given,

    Mass of the body,     m = 3 kg

    Initial speed of the body, u = 2 m/s

    Final speed of the body, v = 3.5 m/s

    Time, t = 25 s

    Using the first equation of motion, 

                         v = u + at

    Acceleration(    a) produced in the body, a =
    Therefore, 



    Now, according to Newton's second law of motion, we have

    F = ma

       = 3 x 0.06

       = 0.018 N

    Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion. 
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    The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. 
    Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

    Final speed of the three-wheeler, v = 0 m/s

    Time, t = 4 s

    Mass of the three-wheeler, m = 400 kg

    Mass of the driver, m' = 65 kg

    Total mass of the system, M = 400 + 65 = 465 kg

    According to the first law of motion,

    Acceleration (    a) of the three-wheeler is, 

                           v = u + at 

    Therefore,

         a = 

    The negative sign indicates that the velocity of the three-wheeler is decreasing with time.

    Now, using Newton’s second law of motion, the net force acting on the three-wheeler is, 

           F = Ma

             = 465 × (–2.5)
     
             = –1162.5 N 

    The negative sign indicates that the force is acting against the direction of motion of the three wheeler. 
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    A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.
     

    Given,

    Mass of the body, m = 0.40 kg

    Initial speed of the body, u = 10 m/s, due north

    Force acting on the body, F = –8.0 N

    Acceleration produced in the body, a =

    (i)

    At     t = –5 s,

    Acceleration, a' = 0

    Initial velocity,     u = 10 m/s 

    Using the relation, 

             s = ut + (1/2) a' t2 

               = 10 × (–5) = –50 m

    (ii)

    At     t = 25 s

    Acceleration, a'' = –20 m/s2 

    Initial velocity, u = 10 m/s

    USing the relation, 

                s' = ut' + (1/2) a" t2

                   = 10 × 25 + (1/2) × (-20) × (25)2 

                   = 250 - 6250

                   = -6000 m

    (iii)

    At     t = 100 s, 

    For 0 ≤ t ≤ 30 s 

    Acceleration, a = -20 ms-2

    Initial velocity, u = 10 m/s

    Now, using the equation of motion, we have

                   s1 = ut + (1/2)a"t2 

                       = 10 × 30 + (1/2) × (-20) × (30)2

                       = 300 - 9000 

                       =  -8700 m

    For 30 < t ≤ 100 s,

    For t= 30 sec, as per the first equation of motion final velocity is given as, 

                       v     = u + at 

                         = 10 + (–20) × 30

                         = –590 m/s 

    Velocity of the body after 30 s = –590 m/s 

    For motion between 30 s to 100 s, i.e., in 70 s: 

    s2 = vt + a" t2 

        = -590 × 70

        = -41300 m 

    ∴ Total distance, s" = s1 + s2 

                              = -8700 -41300

                              = -50000 m
                              = -50 km.
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    A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

    Mass of the body, m = 5 kg

    The given situation can be represented as follows:


    The resultant of two forces is given by, 

    R = 
     is the angle made by R with the force of 8 N.

    Therefore, 


    The negative sign indicates that θ is in the clockwise direction with respect to the force of magnitude 8 N.

    According to Newton’s second law of motion

    Acceleration (    a) of the body is given as, 

    F = ma 

    Therefore, 

    a =

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    A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.
     
    Mass of the rocket, m = 20,000 kg

    Initial acceleration, a = 5 m/s2

    Acceleration due to gravity, g = 10 m/s2

    Using Newton’s second law of motion,

    Net force (thrust) acting on the rocket is given by the relation, 

            F – mg = ma 

                     F = m (g + a)

                        = 20000 × (10 + 5)

                        = 20000 × 15

                        = 3 × 10    5 N
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