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Laws of Motion

Essay type answers Questions

1. State Newton's Second Law of Motion. Hence derive the equation of motion F = ma from it.

    A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body?

    A force of 20 N acts on a body of mass 5 kg at rest. What is the acceleration of the body? What is the velocity after 5 seconds if the same force acts? After 5 seconds if the force ceases to act how will the body move?

Ans: Newton's Second Law of Motion:

Statement: The rate of change of momentum of a body is directly proportional to the resultant or net external force acting on the body and takes place in the direction in which the forces acts.
         If F is the resultant force,
                              
        where ∆p - change in momentum
                    ∆t - change in time
Derivation of F = ma
   Consider a body of mass 'm' moving with a velocity 'v' under the action of a net external force 'F' in the direction of velocity.

   If the velocity is increased by ∆v in a time interval ∆t then by the Second Law 
                                 
  But momentum p = mv

                             
  Assuming the mass of the body m is constant 
                                
  When the time internal ∆t is infinitesimally small
                               
because rate of change of velocity is called acceleration   F = K ma
 Here k is proportionality constant.
The proportionality constant is made equal to one by properly setting the unit of force.
The SI unit of force is newton.
1 newton is defined as the force that causes an acceleration of 1 ms-2 on the body of mass 1 kg substituting these values in F = K.ma, we get
                                               1 = K.1.1 or K = 1.
                   Hence we write F = ma

A body moving in circular path:

  Consider a body moving in a circular path with uniform speed.

  The velocity of the body at any instant is the tangent drawn to it. As the direction of velocity is changing continuously, the body will have acceleration.

  According to Newton's Second Law the direction of net force is in the direction of acceleration.

  This force is called centripetal force acts towards the centre of the circular path of the body.

Problem:

i) Given F = 20 N,   m = 5 kg,   a = ?
                      From F = ma
       

                       
ii) Given u = 0, a = 4 ms-2 (same force) t = 5s,  v = ?
                      using v = u + at
                                v = 0 + 4 × 5 = 20 ms-1
iii) After 5 seconds if the force ceases to act, the body continues to move with uniform velocity 20 ms-1 along straight line. This is by the application of Newton's first law in the absence of friction.

2. Two unequal masses are connected by a very light string over a clamped light smooth pulley. Find the acceleration of the system and the tension in the string.
    Two masses of 3 kg and 4 kg are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Find the acceleration of the masses and the tension in the string, when the masses are released.

Ans: Consider two unequal masses m1 and m2 (m1 > m2)
Let they are connected by a very light string going over a frictionless pulley.
Due to the masses attracted some tension T will be developed in the string.
As m1 is assumed greater than m2 a common acceleration is produced. This is called Atwood's machine.

First body:
  Let us draw the free body diagram for it.
  The forces acting on it are
  i) weight of the body m1g vertically downwards.
 ii) tension in the string T vertically upwards.
iii) the resultant force m1a vertically downwards.
      Hence, m1a = m1g -T ------------------------ (1)    

Second body:
 Let us draw the free body diagram for it
 The forces a acting on it are
  i) weight of the body m2g vertically downwards.
 ii) tension in the string T vertically upwards.
iii) the resultant force m2a vertically downwards.
       Hence, m2a = T - m2g ............................ (2)                   
Adding eq (1) and (2) we get
                            
  This is the common acceleration of the system and its direction is towards heavier mass.
  Substituting the value of 'a' in equation (1), we get
                            
       This gives the tension in the string.

Problem:
            Given m1 = 4 kg , m2 = 3 kg
                     let g = 9.8 ms-2
a) Acceleration
                         
                    a = 1.4 ms-2
b) Tension in the string

Posted Date : 17-11-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

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