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  • A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F. (in new

A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F. (in newtons) 
 

Option: 1

10
 


Option: 2

30


Option: 3

50

 


Option: 4

60


Answers (1)

best_answer

 

 

Newton’s Second and Third Law of motion -

  1. Newton’s second law of motion:-

  • It states that the acceleration of the particle measured from an inertial frame is given by the (vector) sum of all the forces acting on the particle divided by its mass (only when mass is constant), i.e.,

                          \vec{a}=\frac{\vec{F}}{m}\Rightarrow \vec{F}=m\vec{a}

  • Impulse:

  1.  The quantity   \vec{I}=\int_{t_{1}}^{t_{2}}\vec{F}\cdot dtis known as the impulse of the force F during the time interval t1 to t2 and

Is equal to the change in the momentum of the body on which it acts,

i.e. P_f-P_1=\int_{P_1}^{P_1}dP=\int \frac{dP}{dt}*dt=\int Fdt⇒ Area under force and time graph is an impulse.

                                

             2.    Dimension- MLT-1

            3.    Unit- kg-m/sec

  • Impulse Momentum Theorem- Newton’s 2nd law can also be written as:

                      Rate of change in momentum = Force Applied

                           \vec{F}=\frac{d\vec{p}}{dt}

 

 

 

 

Given, m=2 kg, s=10m, t=2s, initial velocity, u=0

s=ut+\frac{1}{2}at^{2}\Rightarrow 10=(0\times t)+\frac{1}{2}\times a\times 2^{2}\Rightarrow a=5 m/s^{2}

F=ma=2\times 5=10 N

Posted by

Gautam harsolia

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