Question

an object weighing 98N is dropped from a height of 10m. it is found to be moving with a velocity 12m/s just before it hits the ground. how large was the frictional force acting upon it?​

Answer 1

Answer:

zero as frictional force is contact force

Answer 2

Given that :

Weight of the object, W = 98 N

It is dropped from a height of 10 m, h = 10 ,

Initial velocity of the object, u = 12 m/s

Final velocity of the object, v = 0 (as it hits the ground)

To find :

The frictional force acting upon it.

Solution :

It can be calculated using Work -energy theorem as :

$fd=\dfrac{1}{2}m(v^2-u^2)$

f is frictional force

m is mass of the object

$m=\dfrac{F}{g}=\dfrac{98}{10}=9.8\ kg$

$f=\dfrac{mu^2}{2d}\\\\f=-\dfrac{9.8\times 12^2}{2\times 10}\\\\f=-70.56\ N$

So, the frictional force acting upon the stone is 70.56 N.

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Work energy theorem

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