An 70 kg man drops from rest on a diving
board −2.9 m above the surface of the water
and comes to rest 0.58 s after reaching the
water.
The acceleration due to gravity is
9.81 m/s
2
.
What force does the water exert on the
man?
Answer in units of N
I think it would just be Fg x m, so 70kg times 9.81 m/s^2, and then multiply all of that by 2.9m, and the negative is just saying what direction the man is falling, meaning downwards
mgh=mv²/2
v=sqrt{2gh} =sqrt{2•9.8•2.9} =7.5 m/s
v(fin)=v-at
v(fin)=0
a=v/t
x=vt -at²/2=vt/2
mv²/2 =W(fr) = F(fr) •x =>
F(fr)= mv²/2x = 2mv²/2 vt=mv/t =
=70•7.5/0.58 = 905.2 N
The above equation is wrong
To find the force that the water exerts on the man, we need to use the principles of Newton's second law and the concept of work and energy.
1. First, let's find the initial velocity of the man when he hits the water. We can use the displacement and time taken provided in the question.
Here, the man drops from rest, so his initial velocity (u) is 0 m/s. The displacement (s) is given as -2.9 m, and the time (t) is given as 0.58 seconds.
We can use the equation: s = ut + (1/2)at^2
Since the man is dropping down, we'll take the downward direction as positive, and the acceleration (a) due to gravity is -9.81 m/s^2.
Plugging the values into the equation, we have: -2.9 = (1/2)(-9.81)(0.58)^2
Solving the equation, we find that the initial velocity (u) of the man when he hits the water is approximately -3.964 m/s.
2. Next, let's calculate the final velocity (v) of the man just before hitting the water. The final velocity is 0 m/s since the man comes to rest after reaching the water.
3. To find the force (F) that the water exerts on the man, we can use the equation: Work (W) = Force (F) * Distance (d)
The work done on the man by the water is equal to the change in kinetic energy of the man while he decelerates in the water.
The work done is given by: W = (1/2)mv^2 - (1/2)mu^2, where m is the mass of the man.
Plugging in the values, we have: W = (1/2)(70)(0^2) - (1/2)(70)(-3.964^2)
Simplifying the equation, we find that the work done is approximately 439.53 J (Joules).
4. Finally, we can find the force (F) exerted by the water on the man using the equation: W = F * d, where d is the distance the man drops.
Plugging in the values, we have: 439.53 = F * 2.9
Solving for the force (F), we find that the water exerts a force of approximately 151.56 N (Newtons) on the man.