An 75 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.6 s after reaching the water.
The acceleration due to gravity is 9.81 m/s2.
What force does the water exert on the man?
Answer in units of N.
To find the force exerted by the water on the man, we can use the equation:
force = mass × acceleration
First, let's calculate the man's initial velocity when he reaches the water's surface.
We can use the equation of motion:
final velocity = initial velocity + (acceleration × time)
Given that the final velocity is 0 m/s (because he comes to rest), the acceleration is 9.81 m/s^2, and the time is 0.6 s, we can solve for the initial velocity:
0 m/s = initial velocity + (9.81 m/s^2 × 0.6 s)
Simplifying the equation:
- (9.81 m/s^2 × 0.6 s) = initial velocity
-5.886 m/s = initial velocity
Now we can calculate the force using the equation:
force = mass × acceleration
Given that the mass is 75 kg and the acceleration is 9.81 m/s^2, we substitute these values into the equation:
force = 75 kg × 9.81 m/s^2
Calculating the force:
force = 735.75 N
Therefore, the force exerted by the water on the man is 735.75 N.
To find the force exerted by the water on the man, we can use Newton's second law of motion, which states that force is equal to mass times acceleration. First, let's find the acceleration of the man.
Given:
Mass of the man (m) = 75 kg
Acceleration due to gravity (g) = 9.81 m/s^2
The initial velocity (u) of the man is zero because he starts from rest.
We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the man comes to rest)
u = initial velocity (0 m/s)
a = acceleration
s = displacement (distance between the diving board and the water)
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0^2 - 0^2) / (2 * (-3.1 m))
a = 0 / (-6.2 m)
a = 0 m/s^2
We find that the acceleration of the man is 0 m/s^2. This indicates that the man is not subject to any horizontal forces during the fall.
Now we can calculate the force exerted by the water on the man using Newton's second law.
F = m * a
Substituting the known values:
F = 75 kg * 0 m/s^2
F = 0 N
Therefore, the force exerted by the water on the man is 0 N.