a)
We would apply one of Newton's formula of motion which is expressed as
s = ut + 1/2at^2
where
s is the distance travelled
u is the initial velocity
a is the acceleration
t is the time
From the information given,
s = 2
t = 1.45
u = 0 since it started from rest
Thus,
2 = 0 * t + 1/2 * a * 1.45^2
2 = 1.05125a
a = 2/1.05125
a = 1.902
The acceleration of the block is 1.902 m/s^2
c) We would calculate the frictional force, f. Taking the positive x axis down parallel to the incline in the direction of the acceleration, we would apply Newton's second law as shown below
Net force = ma
where
m is mass
a is acceleration
Force aiding the motion doen the plane = mgsinθ
frictional force = f
Net force = mgsinθ - f = ma
f = mgsinθ - ma = m(gsinθ - a)
m = 3 kg
θ = 35.5
g = 9.8
f = 3(9.8Sin35.5 - 1.902)
f = 11.37 N
The frictional force is 11.37 N
b) The free body diagram is shown below
N is the normal reaction. Thus,
N = mgcosθ
f = μkN
where
μk is the coefficient of kinetic friction
f is the frictional force
Thus,
μk = f/N = 11.37/(3 x 9.8cos35.5)
μk = 0.475
The coefficient of kinetic friction is 0.475
d) We would apply one of Newton's formula of motion which is expressed as
v = u + at
where
v is the final velocity
Thus,
v = 0 + 1.902 x 1.45
v = 2.76 m/s
The speed of the ball is 2.76 m/s