Question

when a body of mass 10 kg lies on a rough horizontal surface an acceleration of 5 m/s and when the horizontal force us doubled , it gets an acceleration of 18 m/s2(2 is the square). Then the cofficient of froction beetween the body and the horizontal surface is:

A) 0.2 B) 0.8 C) 0.4 D) 0.6

Answer 1

Hello!

The Net force = External Force - Frictional force

this is the force actually exerted on the object. So,

in the first case  F1 = f1 - μmg = ma1 = 50  (1)

in the second case  F2 = f2 - μmg = ma2 = 180  (2)

here

f1 and f2 are external forces in case 1 and 2 respectively, the frictional force = μmg, a1 and a2 are the initial and final accelerations.

here

m = 10 kg

g = 10 m/s

a1 = 5 m/s2

a2 = 18 m/s2 

and the final external force is twice the initial one, so  f2 = 2f1

 

Lets perform the following operation, equation (3) = equation (2) - 2X equation (1)

so,

 180 - 100 = f2 - μmg - 2X(f1 - μmg)

 80 = 2f1 - μmg - 2f1 + 2μmg

or  80 = μmg

thus,     the coefficient of friction μ = 80/(10x10) = 0.8

the answer is option (b)

Answer 2

Answer:

Explanation:

I think this answer will help u

So the answer is 0.8