Question

A constant horizontal force of 40.0 N acts on a body on a smooth horizontal surface. The body starts from rest and is observed to move 100 m in 5.00 s. If the force ceases to act at the end of 5.00 s, how far (meters) will the body move in the next 5.00 s?

Answer 1

Given   : A constant horizontal force of 40.0 N acts on a body on a smooth horizontal surface.

The body starts from rest and is observed to move 100 m in 5.00 s.

the force ceases to act at the end of 5.00 s,

To Find : how far (meters) will the body move in the next 5.00 s

Solution:

S = ut + (1/2)at²

S = 100 m

t = 5 sec

a = ?

u = 0   ( start from rest )

=> 100  = (1/2)a(5)²

=> a = 8  m/s²

V = u + at

=> V = 0 + 8 * 5

=> V = 40 m/s

Velocity after 5 secs  = 40 m/s

force ceases to act at the end of 5.00 s,

as no force so no acceleration

S = Vt

=> S= 40 * 5

=> S = 200  

body move in the next 5.00 s = 200 m

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Answer 2

Question :

A constant horizontal force of 40.0 N acts on a body on a smooth horizontal surface. The body starts from rest and is observed to move 100 m in 5.00 s. If the force ceases to act at the end of 5.00 s, how far (meters) will the body move in the next 5.00 s?

Answer :

$s \: = ut \: + ( \frac{1}{2})at {}^{2} \\ \\ s = 100 \: m \: \\ t = 5 \: sec \: \\ u = 0 \\ a = 100 = ( \frac{1}{2} )a(5) {}^{2}$

a = 8 m/s^ 2

v = u + at

v = 0 + 8 × 5

v = 40 m/s

Velocity after 5 sec = 40 m/s

then

how far (meters) will the body move in the next 5.00 s?

s = v × t

s = 40 × 5

s = 200 m