Question

A stone of 1kg is thrown with a velocity of 50m/s along a surface of earth comes to rest after traveling distance of 100 m . what is the fractional force between stone & earth surface..

Answer 1

u = 50 m/s
v = 0
s= 100 m
using third equation of motion
2*a*s = v^2- u ^2
2* a * 100 = 0- 2500
200 a = -2500
a = -12.5
F = m*a
F = 1* (-12.5)
F = -12.5 N
(-) sign shows opposite direction of force and acceleration

Answer 2

Given ---
Mass of stone (m) = 1 kg
Initial velocity (u) = 50 m/s
Final velocity (v) = 0 m/s....(because the stone comes to rest)
Distance covered (s) = 100 m

To find ---
Frictional force (F)

Finding ---

We know, using the third equation of motion, that

$2as = {v}^{2} - {u}^{2}$

$= > a = \frac{ {v}^{2} - {u}^{2} }{2s}$

$= > a = \frac{ {0}^{2} - {50}^{2} }{2 \times 100}$

$= > a = \frac{0 - 2500}{200}$

$= > a =\frac{ - 2500}{200}$

$= > a = - 12.5$

Hence, acceleration is -12.5 m/s². Negative sign shows that the stone is deacclerating or retarding.

Now, we know that,

$F = ma$
$= > F = 1 \times - 12.5$
$= > F = - 12.5$
Hence, the frictional force between the stone and Earth's surface is -12.5 N.
Negative sign shows that force is acting in the opposite direction to that of motion of stone.

Hope it'll help.. ;-)