Help... conservation of energy...?

Question

You designed a stunt in which a man, who weighs 60 kg, is shot out of a cannon that is elevated 40° from the horizontal. The “cannon” is actually a 0.91-m diameter tube that uses a stiff spring and a puff of smoke rather than an explosive to launch the man. The manual for the cannon states that the spring constant is 1,822 N/m. The spring is compressed by a motor until its free end is level with the bottom of the cannon tube, which is 1.5 m above the ground. A small seat is attached to the free end of the spring for the man to sit on. When the spring is released, it extends 2.7 m up the tube. Neither the seat nor the chair touches the sides of the 3.7-m long tube, such that friction can be neglected. After a drum roll, the spring is released and the man will fly through the air. You have a 0.91-m thick airbag waiting for the man to land on. You need to determine if the airbag is thick enough to stop the man safely – that is, so he is slowed to a stop by the time he reaches ground level

I really need this asap
Please indicate the calculations (w/ explanation pls)

Answer 1

The first step in solving a conservation of energy problem is to do an energy audit at each point where the energy is converted. To do this audit, recognize that E = P + K + W = constant; where E is the total energy of a system (e.g., the man), P is potential energy, K is kinetic, and W = fd is work done on or by the system. f is a net force acting over and in the direction d.

Here are the points of conversion:

Point 0: the starting point with the man at rest in the barrel...E = P(s) + P(z); where P(s) = potential energy of the spring = k dL dL and k = 1,822 kg m/sec^2//m (N/m) and dL = 2.7 m compression/extension of the spring. P(z) = mgz = PE of the man resting in the tube z = 1.5 m above the ground. W = fd = f*0 = 0 here because no work is being done as there is no displacement of the man (d).

Point 1: at the muzzle after the spring is released...E = P(z) + K(m); where K(m) = kinetic energy of the man = 1/2 mv^2 and v = the muzzle velocity P(z) ~ mgz but not quite because the z while at rest is clearly not the height of the muzzle, which is a bit higher off the ground, when the man emerges. But it ought to be close enough. If it isn't for your purpose, add D sin(theta) to z = 1.5 m; where D = length of the barrel to get Z = D sin(theta) + z = height of the man at the muzzle; so that P(Z) = mgZ.

Point 2: apex of the man's trajectory, the highest point...E = P(H) + K(m); where P(H) = mgH = P(h) + P(Z) and H = height above ground = h + Z and h = height above the muzzle elevation (Z). Note Ky(H) = 0 the KE due to vertical velocity Vy = 0 when reaching the top of the trajectory. Also note Kx(H) = 1/2 mVx^2; where Vx is the horizontal velocity parallel to the ground. Vx does not diminish given no drag forces so it remains fixed at the horizontal muzzle velocity. K(m) = Ky(H) + Kx(H) = Kx(H); so E = P(H) + Kx(H) at this point.

Point 3: on impact with the bag of T = .91 m thickness...E = K(T) + P(T); where K(T) = Kx(H) + Ky(T) = kinetic energy at T distance off the ground (impact point). Ky(T) = kinetic energy due to the fall from H to T = 1/2 mUy^2 where Uy is the vertical velocity upon impact. Vx is the horizontal velocity as before. P(T) = mgT, the potential energy, relative to ground level, upon impact.

Point 4: on the ground safe and sound, with the bag deflated (I presume)...at this point P(0) = mg0 = 0 and K(0) = 1/2 m0^2 = 0. As energy can be neither created nor destroyed, that PE > 0 and KE > 0 at impact T height had to be converted into something. And that something was work W = fT = (F - w)T; where w = mg, the weight of the man pulling him to the ground and F = the force of the bag pushing him upwards. f = F - w, the net force acting over the distance T as the bag deflates. Thus, E(T) - W = E(0) = 0, the energy found upon impact at T is converted to work (W), which gives the man no potential or kinetic energy when he and the bag hit the ground.

Now we can backtrack from Point 4 through the other points, to find out if T = .91 is sufficient depth of the bag to give the man E = 0 back on the ground. Thus, we can write E(T) = P(T) + K(T) - (F - w)T = 0; so that, P(T) + K(T) = (F - w)T = E(T)

Then we solve for T, the thickness needed to convert all the PE and KE at impact to zero at ground level. T = E(T)/(F - w) = E(T)/m(a - g) as w = mg and F = ma. It is clear that the bag has to provide sufficient force to make the deceleration a greater than g; thus, a > g must be true to stop the man. The issue is, just how big a must be to stop him just at ground level.

So we find a, when T = .91 m. We have maT - mgT = E(T); then aT = [E(T) + mgT]/m = E(T)/m + gT and a = E(T)/mT + g. This is good news, after all this the last term at least is an acceleration term; so it appears our units are consistent with a. Now we need to find E(T) in known variables; so we can solve for a.

From our audit we see that E(T) = P(T) + K(T); P(T) = mgT and they are all givens...good start. K(T) = Ky(T) + Kx(T), the KE from falling to T from H, the high point in the trajectory, and from the Vx velocity out of the muzzle. This might be problematic, but for one thing...the conservation of energy law.

Note this Kx(T) = P(s) cos(theta); that is, the horizontal velocity and KE comes from converting the horizontal component of the spring's PE into that kinetic energy. And P(s) = k dL^2 and theta = 40 deg; so all the variables going into Kx(T) = k dL^2 cos(theta) are also known. We're getting there. Last thing, Ky(T).

This one's not easy...we have multiple Y lengths involved in the potential energies and we still have the Y component of the spring's PE= Py(s) = P(s) sin(theta) = Ky(h) where h = H - Z the distance traveled from the muzzle to the top of the trajectory.

But we need Ky(d), which is the KE after falling a distance d from H to T, not Ky(h) which is the KE after falling h. So here we note that the fall distance to T from H is just d = H - T = h + Z - T. Thus the PE for a drop of d = Py(d) = Py(H) - Py(T) = Py(h) + Py(Z) - Py(T) = P(s) sin(theta) + mgZ - mgT = k dL^2 sin(theta) + mg(Z - T) = Ky(d). This follows from the conservation of energy. Now we have K(T) = Ky(d) + Kx(T) = P(s) sin(theta) + mg(Z - T) + P(s) cos(theta) and P(T) = mgT, all in known variables.

Putting it all together a = E(T)/mT + g = [P(T) + K(T)]/mT + g = [mgT + k dL^2(sin(theta) + cos(theta)) + mg(Z - T)]/mT + g = [k dL^2 (sin(theta) + cos(theta)) + mgZ]/mT + g. This give you the a necessary to stop the man exactly at ground level with zero excess energy.

The issue now is...what force does that man endure when decelerated at a m/sec^2? We know this, a > g from the derived equation above. This means that f = ma > mg = the weight of the man. Thus the force to stop him in time is more than about mg = 60*10 ~ 600 Newtons (g ~ 10 m/sec^2). You need to do the math, all the variables in the derived equation are given; so just plug in the numbers. I suggest that if f > than that from impacting the ground at, say, 120 mph, then the a is too high and the poor guy will not survive.

Noteworthy is that a = [k dL^2 (sin(theta) + cos(theta)) + mgZ]/mT + g = Py(s) + Px(s) + P(Z)]/mT + g; so that maT = P(s) + P(Z) + mgT and m(a - g)T = (F - w)T = P(s) + P(Z) = W. In other words, as we might have guessed outright, the bag has to do enough work to dissipate the potential energy of the spring P(s) and of the man's potential at the muzzle P(Z), which is all the potential just prior to firing.

Despite what one answerer indicated, I think this was an excellent question...worthy of advanced AP or Freshman college physics.

PS: After working through the audit, it is clear that the fastest way to work this problem is to recognize that Total PE in the barrel = Work of the bag. That is, as long as there is no other work, like from air drag or barrel friction, the air bag has to come up with zero E, total energy, when the man hits the ground in the bag. And the only way that can happen is to convert all that PE, which is total energy at Point 0, while in the barrel to zero; so that we have E = P(0) = P(s) + P(A) = m(a - g)T.

If there is air or barrel friction, some of the P(0) will be converted to frictional energy; so the bag will have less work to do to bring the man to a halt. This follows from P(0) - friction energy = W - friction energy = w net work of the bag after friction energy is removed from the potential energy in the barrel. Unfortunately, friction energy through the air depends on relative velocity of the man in flight. And as relative velocity is in constant change over the trajectory, this is not easy to calculate.

Answer 2

Where from, do you expect it to come ? What we know about energy is, 'it exists around us' There is always some change in the forms of energy. The total remains the same. When two persons are having a total of 100 dollars, what ever the way they share these amounts, the total will remain as 100 dollars. This is law of conservation of energy . Your question is how did they get 100 dollars. The law cannot answer this question. It is a statement about sharing the money and not the source of income for the two. As long as they do not interact with others the 100 dollars will not change. They cannot create dollars or loose dollars as long as they are there together. Now if another person joins with these two with another 50 dollars, then the total will now become 150 dollars and the law is now about the total of 150 dollars and not about 100 dollars. The law do not question how did the third person get 50 dollars. If you include all world and all nations the total amount of money will remain the same. If you include persons who have gone to mars or other planets, again the total amount will remain the same.

Answer 3

This is an awful quesiton and you need to slap your physics teacher right in the mouth for making you do it. Knowing that the airbag is .91m thick is of aboslutely no use whatsoever unless you also know the PSI inside it. Unless of course there is no pressure and they want you to determine if the material itself is enough of a cushion, in which case you still need to know some of its properties like resilience and compression rate. And really, look at the wording of the last sentence. Yes, he will be slowed to a stop by the time he reaches ground level, he would have been even without an airbag. Things stop when they hit the ground (or bounce, either way he's probably going to die). There are a number of other mistakes with this question that I won't take further space pointing out, but suffice it to say any answer you can give with the information stated in that question is no better than a guess. I hope there is further reference material to this question, exact numbers of USEFUL information. Otherwise, ask your teacher if the class can study Dianetics next since he's apparently into basing calculated answers on sketchy or nonexistant data. FYI, the law of conservation of energy simply states that the total amount of energy in an isolated system remains constant, so he will be hitting the airbag with the same amount of energy as was applied to him by the cannon (since we are neglecting friction here, which would severely lessen his force due to wind and mechanical friction changing that kinetic energy into thermal). I don't know how that helps honestly, but it was in the question so i figured i'd include it.

Answer 4

You need to ask Steven Hawkins that ?