Question

a constant force an object of mass 5kg for a duration of 2s it increase the objects velocity from 3m/s to 7m/s find the magnitude of the applied force .now if the forces was applied for a duration of 5s what would be final velocity of object​

Answer 1

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.

Answer 2

• initial velocity=u=3m/s
• Final velocity=v=7m/s
• Time=t=2s
• Mass=5kg=m

Using newtons second law

$\boxed{\sf F=ma}$

$\\ \sf\longmapsto F=m\dfrac{v-u}{t}$

$\\ \sf\longmapsto F=\dfrac{5(7-3)}{2}$

$\\ \sf\longmapsto F=\dfrac{5(4)}{2}$

$\\ \sf\longmapsto F=5\times 2$

$\\ \sf\longmapsto F=10N$

Using first equation of kinematics

$\boxed{\sf v'=u+at}$

$\\ \sf\longmapsto v'=3+2(5)$

$\\ \sf\longmapsto v'=3+10$

$\\ \sf\longmapsto v'=13m/s$