A 75.1 kg diver jumps from a 10.0 m platform into the water. After hitting the water, the diver submerges 2.19 m until they stop. What is the net force (kN or kiloNewtons) acting on the of the diver after hitting the water?
To find the net force acting on the diver after hitting the water, we can use the principles of mechanics.
First, we need to determine the initial velocity of the diver just before impact. We can use the equation for free fall motion:
v^2 = u^2 + 2as,
where v is the final velocity (0 m/s because the diver stops after submerging), u is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (10.0 m platform height).
Rearranging the equation, we have:
u^2 = v^2 - 2as,
u^2 = 0^2 - 2(-9.8 m/s^2)(10.0 m),
u^2 = 196 m^2/s^2,
u = √(196) m/s,
u = 14 m/s.
The initial velocity (u) of the diver just before impact is 14 m/s.
Next, we can use the equation of motion to calculate the stopping time of the diver in the water:
v = u + at,
where v is the final velocity (0 m/s), u is the initial velocity (14 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the stopping time.
Rearranging the equation, we have:
t = (v - u) / a,
t = (0 m/s - 14 m/s) / (-9.8 m/s^2),
t = -14 m/s / -9.8 m/s^2,
t = 1.43 s.
The stopping time (t) is 1.43 seconds.
Now, we can calculate the deceleration of the diver using the equation:
a = Δv / Δt,
where a is the acceleration, Δv is the change in velocity (final velocity - initial velocity), and Δt is the change in time (stopping time).
Rearranging the equation, we have:
Δv = a * Δt,
a = Δv / Δt,
a = (0 m/s - 14 m/s) / (1.43 s),
a = -9.79 m/s^2.
The deceleration (a) of the diver in the water is -9.79 m/s^2.
Finally, we can find the net force acting on the diver after hitting the water using Newton's second law:
F = ma,
where F is the net force, m is the mass of the diver (75.1 kg), and a is the deceleration.
Substituting the values, we have:
F = (75.1 kg) * (-9.79 m/s^2),
F ≈ -734.729 N.
The net force acting on the diver after hitting the water is approximately -734.729 N (negative because it acts in the opposite direction of the diver's motion).