A child of mass 50kg rides on 'upih' sled which restd on a sand-covered hill making an inclined of theta

with the horizontal. the coefficient of static friction is 0.34. determine,
a) maximum angle of inclination of the hill at which the sled is at rest
b) the coefficient of kinetic friction between the sand and the upih when a force is applied and the upih slides with an acceleration of 1.05m/s2

A) mgsin(theta) = f1
mgsin(theta) = umgcos(theta)
tan(theta) = u
tan(theta) = 0.34
theta = 18.8degree

B) a= 1.05m/s , let F is the force , m = 50kg
f = uN = umgcos(theta)
mgsin(theta) - f = ma
50gsin(theta) - umgcos(theta) = 50*1.05
50g(sin(18.8) - u*cos(18.8)) = 50*1.05
g*(0.322 - u*0.94) = 1.05
0.322 - u*0.94* = 0.107
0.322 - 0.107 = u*0.94
u = 0.2

3x(2)c - 2c(5) +5c - 3 28x + 30
________________________ = (3x - 11) + _______________

x(2) + 3x + 3 x(2) + 3x + 3

The answer is 34.4 m/s(2)