Answer:
28716.4740661 N
1.2131147541 m/s
51.2474965841%
Explanation:
m = Mass of plane = 74000 kg
s = Displacement = 3.7 m
f = Frictional force = 14000 N
t = Time taken = 6.1 s
u = Initial velocity = 0
v = Final velocity

Force is given by

The force with which the team pulls the plane is 28716.4740661 N

The speed of the plane is 1.2131147541 m/s
Kinetic energy is given by

Work done is given by

The fraction is given by

The teams 51.2474965841% of the work goes to kinetic energy of the plane.
Answer:
Part a)

Part b)

So this speed is independent of the mass of the rider
Explanation:
Part a)
By force equation on the rider at the position of the hump we can say

now we will have


now we have



Part b)
At the top of the loop if the minimum speed is required so that it remains in contact so we will have

at minimum speed




So this speed is independent of the mass of the rider
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s