Math, asked by ravenclaw61, 3 months ago

Solve the attached pic plzz​

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Answered by SweetCharm
29

Given :-

Mass of object,m= 5 kg

Time taken by object,t= 2s

Initial velocity of object,u= 3 m/s

Final velocity of object,v= 7 m/s

To Find :-

The magnitude of the applied force and the velocity of object if the force is applied for a duration of 5s

Solution :-

We know that,

Acceleration of a particle 'a' is given by

\pink{\underline{\boxed{\bf{a=\frac{v-u}{t}}}}}

Expression of force F is given by

\purple{\underline{\boxed{\bf{F=ma}}}}

where,

v is final velocity

u is initial velocity

t is time taken

m is mass

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Let the acceleration of object be a

So,

\longrightarrow\rm{a=\dfrac{v-u}{t}}

\longrightarrow\rm{a=\dfrac{7-3}{2}}

\longrightarrow\rm{a=\dfrac{4}{2}}

\longrightarrow\rm{a=2\ m/s^{2}}

Let the applied force be F

So,

\longrightarrow\rm{F=m\times a}

\longrightarrow\rm{F=5\times 2}

\longrightarrow\rm\green{F=10\ N}

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Let the final velocity of second case be v'

So,

\longrightarrow\rm{a=\dfrac{v'-u}{t}}

\longrightarrow\rm{2=\dfrac{v'-3}{5}}

\longrightarrow\rm{10=v'-3}

\longrightarrow\rm{v'-3=10}

\longrightarrow\rm\green{v'=13\ m/s}

Hence, the applied force is 10 N and the velocity of object if the force is applied for a duration of 5s  is 13 m/s.

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Answered by vanshika2056
1

Answer:

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