Question

A constant horizontal force of 10N acts on a body on a smooth horizontal plane. The boday starts from rest and is observed to move 250m in 5s. If the force ceases to act at the end of 5s, how far will the body move in the next 5s?

Answer 1

S = 0.5at^2
250 = 0.5a * (5)^2
a = 250 * 2 / 25
a = 20 m/s^2

v = u + at
v = 0 + (20 * 5)
v = 100 m/s

If Force ceases to act after 5s then it will move with uniform velocity of 100 m/s continuously until an external force stops it, as it is moving on SMOOTH surface.

For next 5s it move by
S = vT
= 100 * 5
= 500 m

It will 500 m in next 5s

Answer 2

Answer:

the body move 500 m in the next 5s

Explanation:

S = 0.5at^2

250 = 0.5a * (5)^2

a = 250 * 2 / 25

a = 20 m/s^2  

v = u + at

v = 0 + (20 * 5)

v = 100 m/s

For next 5s it move by

S = vT

= 100 * 5

= 500 m

the body move 500 m in the next 5s

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