Question

a constant force acts on the object of mass 5 kg for 2 seconds it increases the object velocity 3 metre per second to 7 metre per second what is the magnitude​

Answer 1

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A constant force acts on an object of mass 5 kg for a duration of 2 seconds .It increases the objects velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force . Now if the tax were applied for a duration of 5 seconds , what would be the final velocity of the object?

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ANSWER

Given,

mass=5kg

t

1

=2s

Initial velocity u=3m/s

Final velocity v=7m/s

t

2

=5s

So,

Let the Force be F

Let the acceleration be a

So,

a=

t

(v−u)

=

2

(7−3)

=2m/s

2

So the magnitude of the applied force is 10N

And the final velocity after 5s is v

So,

v=u+at

v=3+2×5

v=13m/s

The final velocity after 5s is 13m/s

Answer 2

Answer:

$Given,</p><p></p><p></p><p>mass=5kg</p><p></p><p></p><p>t1=2s</p><p></p><p></p><p>Initial velocity u=3m/s</p><p></p><p></p><p>Final velocity v=7m/s</p><p></p><p></p><p>t2=5s</p><p></p><p></p><p>So,</p><p></p><p></p><p>Let the Force be F</p><p></p><p></p><p>Let the acceleration be a</p><p></p><p></p><p>So,</p><p></p><p></p><p> a=t(v−u)=2(7−3)=2m/s2</p><p></p><p>  </p><p></p><p>So the magnitude of the applied force is 10N</p><p></p><p></p><p>And the final velocity after 5s is v </p><p></p><p></p><p>So,</p><p></p><p></p><p>v=u+at</p><p></p><p></p><p>v=3+2×5</p><p></p><p></p><p>v=13m/s</p><p></p><p></p><p>The final velocity after 5s is 13m/s </p><p></p><p>Answered By</p><p></p><p>$