HELP, cant figure out this problem with. A fall item and ave to find the speed when it his the ground ....?

Question

A .20 kg pinecone falls from a branch 18m above the ground.
(a) With what speed would it hit the ground if air resistance could be ignored?
(b) if it actually hits the ground with a speed of 10 m/s, what was the average force of air resistance exerted on it?

Answer 1

Is it 20 kg or .20 kg....anyways....I'm doing for .20 kg

a)
You have the formula : v^2 - u^2 = 2gh ....Equation [1]
u = 0 (no initial velocity).
g = 9.81 m/s^2 ( acceleration due to gravity)
h = 18m

Therfore,
V^2 - 0 = 2*9.81*18
So, v = 18.79 m/s (the speed with which it would hit the ground)

b)
By Newton's Second Law,
F1 = m.g = .2 * 9.81 = 1.962
Again,
F2 = m.a
From equation [1], we get a = 2.778 m/s^2, taking v=10m/s and h = 18m
So,
F = .2 * 2.778
Therefore, the air resistance force is : 1.962 - .5556 N = 1.406 N

Answer 2

v = √(2gh) so
v = √(2*9.8*18) = √352.8 = 18.782 m/s
Now, gravity (9.8m/s²) is an acceleration and, when it acts on a .2 kg pinecone, it generates a force of
F = ma = .2*9.8 = 1.960 N. The question is how much acceleration (average) is required to get the pinecone up to 10 m/s after an 10 m fall? So
v = √(2gh)
v² = 2gh
g = v²/2h = 10²/2*18 = 100/36 = 2.777 which would generate a force of
F=ma=.2*2.777 = .555 N so the force (average) required from air friction must be
1.960 - .555 = 1.405 N to make the pinecone hit the ground at 10 m/s.

HTH

Doug

Answer 3

This is a conservation of energy problem (potential and kinetic energy). I'm sure you remember these formulas:
U = mgh where g is the 9.8 m/s/s (gravity) and h is distance above ground or reference point.
K = 1/2*mv^2 where v is linear velocity in meters/s
m represents mass in both equations.

Thus if you drop a 0.20 kg pinecone from 18 m above ground it will, before you release it, hold 0.20*18*9.8 = 35.28 Joules of potential energy. The instant before it hits the ground it has a speed (really velocity,) of v = sqrt(35.28/0.5/0.20) = approx. 18.8 m/s. If you pay attention to significant figures the answer should be rounded up to 20 m/s.

Part (b): if E = F*d, then F = E/d. If the pinecone has only 10 m/s instead of 18.8 m/s then the deficit in energy is: 35.28 J - 0.5*0.20*10*10 J = 25.28 J
25.28 J / 18 m = 1.4 Newtons average air drag throughout the descent.

Answer 4

Have you ever heard of the old experiment where you take a feather and a quarter and drop them, timing how long it takes for them to land? If it is done in a vacuum tube, then they both land at the same time. My point is the weight of the pinecone is filler that teachers use to distract students with. hehe =)

9.8 meters per second is earths gravitational constant. That means that we're all hurtling towards the earth's core at that speed =P So with no air resistance, and nothing to boost the speed of your lovely pinecone, I think it is safe to say that its speed would be 9.8 meters per second at the moment before it impacted the ground.

If that were somehow sped up to 10 m/s...well I suppose that implies giving your pinecone a tailwind lol, and converting that into newtons.

Good luck with that last one lol.

Answer 5

(1/2)mv^2 = mgh
v = √(2*9.80665*18)
v = 18.789 m/s

F*h = 2*9.80665*18 - (1/2)10^2
F(avg) = 16.836 N