m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
Answer:
1. 60 m/s.
2. 3600 m.
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Time (t) = 2 mins
Final Velocity (v) =?
Distance travelled (s) =?
1. Determination of the velocity at the end of 2 minutes.
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Time (t) = 2 mins = 2 x 60 = 120 secs
Final Velocity (v) =?
v = u + at
v = 0 + (0.5 x 120)
v = 60 m/s
Therefore, the velocity at the end of 2 minutes is 60 m/s.
2. Determination of the distance travelled.
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Final velocity (v) = 60 m/s
Distance travelled (s) =..?
v² = u² + 2as
60² = 0 + 2 x 0.5 x s
3600 = 1 x s
s = 3600 m
Therefore, the distance travelled is 3600 m.