Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is

Answer: Option D: 5.5×10²Joules
Explanation:
Work done is the product of applied force and displacement of the object in the direction of force.
W = F.s = F s cosθ
It is given that the force applied is, F = 55 N
The displacement in the direction of force, s = 10 m
The angle between force and displacement, θ = 0°
Thus, work done on the object:
W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J
Hence, the correct option is D.
Answer:
23.52092 J
Explanation:
m = Mass of block = 6.79 kg
s = Sliding distance = 2.82 m
= Angle of slide = 20.7°
= Coefficient of kinetic friction = 0.425
g = Acceleration due to gravity = 9.8 m/s²
Work done by the force of gravity is given by

The work done by the force of gravity is 23.52092 J