Question

A horizontal force applied on a body on a rough surface produces an acceleration



a. If the coefficient of friction between the body and surface which is m is reduced to m/3 the acceleration increases by 2 units. The value of m is

Answer 1

As per given condition, a2=2+a1; μ1=m; μ2=m/3

Normal reaction = R = mg

frictional force = f = μR = μmg

FBD in first case,

F - f1 = ma1

F - μ1(mg) = ma1

F - m²g = ma1

F = m²g+ma1 ...(i)

FBD in second case,

F - f2 = ma2

F - μ2(mg) = m(2+a1)

F - m²g/3 = 2m + ma1

F = m²g/3+2m+ma1 ...(ii)

From (i) & (ii) we get,

m²g+ma = m²g/3 + 2m + ma

3m²g+3ma = m²g+6m+3ma

3m²g = m²g+6m

2m²g = 6m

m²g = 3m

mg = 3

m = 3/g = 3/10

m = 0.3