Question

a constant force acts on an object of mass 7kg for a duration of 2 second. it increase the velocity of the object from 3m/s to 7m/s . find the value of the applied force

Answer 1

Answer:

t= 2sec

m=7kg

v=4m/s(7-3m/s)

f=ma

F=m×v/t(A=V/t)

F=7×4/2N

F=7×2N

F=14N

Answer 2

Given :-

A constant force acts on an object of mass 7 kg for a duration of 2 seconds .

The velocity of the object had increased from 3 m/s to 7 m/s

Required to find :-

• Value of the applied force ?

Formulae used :-

$\boxed{ \text {\pink{ Acceleration =}} \tt\dfrac{v - u}{t} }$

$\boxed{ \text{ \pink{Force = \:}} \rm{mass \times acceleration }}$

Solution :-

Given information :-

A constant force acts on an object of mass 7 kg for a duration of 2 seconds .

The velocity of the object had increased from 3 m/s to 7 m/s

we need to find the value of the applied force .

From the given information we can conclude that ;

• Mass of the object ( m ) = 7 kg

• Initial velocity of the object ( u ) = 3 m/s

• Final velocity of the object ( v ) = 7 m/s

• Time ( t ) = 2 seconds

In order to find the value of applied force we should first find the value of acceleration .

So,

Using the formula ;

$\tt{ Acceleration = \dfrac{ v - u }{t }}$

$\tt{ Acceleration = \dfrac{ 7 - 3 }{ 2 }}$

$\tt{ Acceleration = \dfrac{ 4 }{2 }}$

$\tt{\bf{\underline{Acceleration = 2 \; m/{s}^{2} }}}$

Hence,

• Acceleration of the object = 2 m/s²

Using the 2nd formula let's find the force applied

$\rm{Force = mass \times acceleration }$

By substituting the values ;

$\rm{ Force = 7 \: kg \times 2 \; m/s^2 }$

$\rm{ Force = 14 \; kg*m/s^2 }$

$\boxed{\because{ 1 \; kg*m/s^2 = 1 \; N }}$

So,

$\implies{\tt{\bf{\underline{ Force = 14 \; N }}}}$

Therefore ,

• Value of the applied force = 14 N

Point to remember :-

>> If we want to solve this numerical in a single strike then use the below formula ;

$\boxed{\rm{\bf{ Force = Mass \times \dfrac{ v - u }{t } }}}$

Because , acceleration = v - u/ t