Science, asked by Akankshagujjar, 1 year ago

a body of mass 100 gram is at rest on the smooth surface of force of 0.2 Newton acts on it for 5 seconds calculate the distance travelled by the body​

Answers

Answered by letshelpothers9
33

Explanation:

mass (m) = 100 g = 0.1 kg

force (f) = 0.2 Newton

time taken (t) = 5 seconds

initial velocity (u) = 0

to find : distance travelled by the body (s)

solution :

force = mass × acceleration

0.2 = 0.1×a

a = 2 m/s^2

by using 2nd equation of motion

s = ut+1/2at^2

s = 0+1/2*2*(5)^2

s = 25 meter

So, the body will travel a distance of 25 meter.


Akankshagujjar: thank you
letshelpothers9: welcome :)
Answered by Anonymous
39

\bf{\large{\underline{\underline{Answer:-}}}}

\Large{\boxed{\sf{s = 25 \: m}}}

\bf{\large{\underline{\underline{Explanation:-}}}}

Given :-

Mass of a body = 100 grams

Force applied on the body when it is at rest = 0.2 Newton

To find :- Distance travelled in 5 seconds

Solution :-

Mass of a body = 100 grams

Force applied on the body when it is at rest = 0.2 Newton

Find acceleration of the body

To find it first see that mass and force are of same unit systems

Here Force - SI

Mass - MKS

So, we will convert Mass units from MKS to SI

SI unit of Mass = Kg

So we need to convert g to Kg

We know that. 1 Kg = 1000 g

So, 1g = 1/1000 Kg

Mass in( SI unit system) = 100 * 1/1000 = 100/1000 = 1/10 Kg

Mass = 1/10 Kg

\boxed{\sf{ \star \: \: Force = Mass \times Acceleration}}

\tt{\implies{0.2 =  \dfrac{1}{10} \times Acceleration}}

\tt{\implies{0.2 =  \dfrac{1}{10} \times Acceleration}}

\tt{\implies{ \dfrac{2}{10} \div  \dfrac{1}{10} =  Acceleration}}

\tt{\implies{ \dfrac{2}{10} \times 10 =  Acceleration}}

\tt{Acceleration=2}

So, Acceleration = 2 m/s²

From above we get information as,

Acceleration(a) = 2 m/s²

Given body started from rest

Initial velocity (u) = 0 m/s

Time(t) = 5 seconds

Distance (s) = ?

These all values are in second equation of motion i.e, s = ut + 1/2 at²

By substituting the values in the equation,

\tt{\implies{s = 0(5) +  \dfrac{1}{2} \times 2 \times  {5}^{2} }}

\tt{\implies{s = \dfrac{1}{2} \times 2 \times 25 }}

\tt{\implies{s = 25 }}

So ,

\Large{\boxed{\sf{s = 25 \: m}}}

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