Question

of the value of g suddenly becomes twice it's value ,it will become two times more difficult to pull a heavy object along the floor.why?​

Answer 1

As you might be knowing,
F = -(G m1 m2)/r²

where F is the gravitational force
G is the gravitational constant
m1 is the mass of 1st body or object
m2 is the mass of 2nd body or object
r of the distance between the 2 masses

Now, this formula is simplified as
F = mg
g → acceleration due to gravity

Now, on this case, if g is doubled, there is no change in mass of the body or object. Because mass is constant, and doesn't change. So F is directly proportional to "g".

$F \: \alpha \: g$

Now if the value of"g" of doubled, then The cake value of F will also be doubled (considering ideal case), since they are director proportional.
So, more force will be needed.

$hope \: you \: understand \: : )$

Thankyou!!!

Answer 2

hello ☺️☺️

initial value of g is = g
suddenly it becomes 2g

For pulling an object along the floor we need to apply force which is greater than frictional force.

Consider a case in which we are pulling a body of mass 'm' towards the right direction. Frictional force will oppose relative motion and will act towards the left direction.

value of frictional force is = u×N
where u is coefficient of frictional force and N is normal reaction. N is equal to the weight

intially value of g was 'g'.... suddenly it changed to '2g'
so initially weight of body was mg... now it changes to 2mg.

so now value of N is 2mg. frictional force acting is u×2mg...( earlier it was u×mg)

hence frictional force doubles. to pull this object we have to apply double the force.

see diagram for reference.

hope this helps you.