Answer:
a) a=1.64 m/s² : magnitude of the acceleration of the block
b) μk = 0.384 : coefficient of kinetic friction
c) Ff = 13.04 N : in direction (-x), forming a 30° angle with the horizontal.
d) v= 2.62 m/s : speed of the block after it has slid 2.10 m.
Explanation:
kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (1)
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
a) v₀=0 ,d= 2.1 m , t= 1.6 s , a=?
We apply the formula (1)
d= v₀t+ (1/2)*a*t²
2.1= 0 + (1/2)*a*(1.6)²
2.1= 1.28*a
a= (2.1)/(1.28)
a= 1,64 m/s²
b)Kinetics analysis
We apply Newton's second law:
∑F = m*a (Formula 3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=4 kg
β = 30°
g = 9.8m/s²
W= m*g = 4 kg*9.8m/s²= 39.2 N : Block weight
x-y weight components
Wx= Wsinβ=39.2 N*sin30° = 19.6 N
Wy= Wcosβ=39.2 N*cos30° =33.95 N
Look at the attached graphic
We apply the formula (3) to calculate the frictional force (Ff)
∑Fx = m*a
Wx-Ff=m*a
Wx-m*a= Ff
Ff = 19.6 - 4* 1.64
Ff =13.04 N
We apply the formula (3) to calculate theNormal force (N)
∑Fy = m*a, a =0 in direction y
N-Wy = 0
N = Wy = 33.95 N
b)Friction force formula
Ff =μk* N
μk = (Ff) / (N)
μk = (13.04) / (33.95)
μk = 0.384
c) Ff =μk* N
Ff = 0.384*33.95= 13.04 N : in direction (-x), forming a 30° angle with the horizontal.
d) Calculation of the final speed at 2.1 m
We apply formula (2) :
vf²=v₀²+2*a*d
[tex]v_{f} = \sqrt{2*a*d} = \sqrt{2*1.64*2.1} = 2.62 m/s[/tex]
