Question

An object of mass 5 kg is kept at rest. A force of 20 N acts on it for 5 s. Find the distance travelled by object in 10 s?

Answer 1

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GIVEN,
◆ F = 20 N
◆ m = 5 kg
◆ u = 0 m/s
◆ t = 5 s
Therefore, Acceleration produced (a)
= 20/5 m/s²
= 4 m/s²




Now Distance covered in this 5 second is given by :
S = ut + 1/2 at²
= (0×5)+{1/2 × 4 × 25} m
= (0+50) m
= 50 m.


So, in 5 second, the distance covered is 50 m.
In 10 second, the distance covered is {(50/5)×10} m
= 100 m.




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Answer 2

Distance travelled by object in 10s is 100 m

Given:

An object of mass 5 kg is kept at rest. A force of 20 N acts on it for 5 s

To Find:

To get the value of the distance travelled by object in 10s

Solution:

Acceleration produced

= 20/5 m/s²

= 4 m/s²

Now Distance covered in this 5 second is given by :

S = ut + 1/2 at²

= (0×5)+{1/2 × 4 × 25} m

= (0+50) m

= 50 m.

So, in 5 second, the distance covered is 50 m

In 10 second, the distance covered is {(50/5)×10} m

= 100 m

Distance travelled by object in 10s is 100 m

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