initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
here we know that
a = 9.8 m/s^2
t = 1.5 s
now from above equation
so it will hit the floor with speed 14.7 m/s
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
