A 65 kg diver jumps off of a 10.0 m tower.
a. Find the diver's velocity when he hits the water.
b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.
Ok, im confused, I found velocity which is 14. But for the net force im confused on what we are finding when using this equation below.
For the force, it would be similar to the car acceleration problem, except that to find the deceleration in water, we would use
v1²-v²=2aS
S=distance (2 m)
v1=final velocity =0
v=initial velocity as found above
a=deceleration due to water (negative) including effects of buoyancy and gravity.
and finally
F=ma
You're quite right, the deceleration as calculated above accounts for gravity, but the force does not.
From v=14, v1=0, we calculate a=(14²-0²)/2/2m=49
F=ma=65*49=3185 N
On this we have to add the weight of the diver, 65*9.8 to give 3822 N.
As suggested by Drwls, it is simpler to do this all in one step by energy considerations, if that's the topic you are working on in class.
Σ PE+KE at start
= Σ PE+KE+work done
PE=potential energy
KE=kinetic energy
At start:
PE = mgh (h=12 m relative to final position)
KE = 0
At end:
PE = 0
KE = 0
Work done = F*2 metres
Therefore
mgh+0 = 0 + 0 + 2F
F=mgh/2
= 65*9.8*12/2
= 3822 N
bob
Gonna have to agree with sbob on this one folks
To solve part a of the problem, you correctly found the diver's velocity when he hits the water as 14 m/s.
Now let's focus on part b, where we need to find the net force exerted by the water on the diver when he comes to a stop 2.0 m below the surface.
To calculate the net force, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the object is the diver and we need to find the net force exerted by the water.
Here's how we can approach the problem step by step:
1. Convert the distance submerged, 2.0 m, into a negative displacement since it is below the surface. So, S = -2.0 m.
2. Use the equation v1² - v² = 2aS and plug in the given values:
- v1 = final velocity = 0 m/s (since the diver comes to a stop)
- v = initial velocity = 14 m/s (as found in part a)
- S = displacement = -2.0 m (negative because it is below the surface)
Let's calculate the deceleration, a:
v1² - v² = 2aS
0 - (14)² = 2a(-2)
0 - 196 = -4a
-196 = -4a
Now, let's solve for a:
a = -196 / -4
a = 49 m/s²
3. Now we have the deceleration, a, which includes the effects of buoyancy and gravity.
4. Finally, we can find the net force, F, by multiplying the diver's mass, m, by the deceleration, a:
F = m * a
Given that the diver's mass is 65 kg, we can calculate the net force:
F = 65 kg * 49 m/s²
F = 3,185 N
So, the net force exerted by the water on the diver is 3,185 Newtons.