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Conservation of energy of a dart gun

  • Thread starter Electro
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  • #1
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Hi!
I have a two-part problem to solve. I think I solved it but in the class the professor told us that there's something wrong with the exercise.

The spring constant of a toy dart gun is 1350 N/m. To cock the gun the spring is compressed 1 cm (0.01 m). The 5g (0.005 kg) dart, fired straight upward, reaches a maximum height of 24 m. g is 9.81 m/s^2. Determine the energy dissipated by air friction during the dart's ascent.

My solution:

E = -1/2 *(k) *(x)^2 + m*g*h
=-0.5*1350* (0.01)^2 + 0.005*9.8*24
= -0.0675 + 1.176 = 1.1085 J
Part 2.

What speed should the projectile have when it returns to its starting point?

Solution: m*g*h = 0.5*m*v^2
v = sqrt(2*g*h)
=sqrt(470.4) = 21.69 m/s

These make sense to me, but i don't know why the professor told me there's something wrong in the problems.
 

Answers and Replies

  • #2
arildno
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2) is inconsistent with 1) since you suddenly ignore air resistance..
might this be what your prof. disliked?
I would agree with him!
 
  • #3
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For part A, the spring stores a certain amount of potential energy. When the gun is fired, the spring transfer all its potential energy to the bullet in the form of kinetic energy. As the bullet travels upwards this kinetic energy converts into gravitational potential energy. Do you see where the problem is?
 
  • #4
arildno
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Eeh, it seems that air friction has ADDED energy to the system, when I look a bit closer.
this exercise is a complete mess.
 
  • #5
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Although i think your reasoning for part B would be the correct one. I think you are supposed to ignore air friction on the way down because it says what speed *should* the bullet have.
 
  • #6
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:smile: In both cases the air resistance is neglected. He said that the exercise doesn't make sense because the numbers come negative. This doesn't happen when I solve the problems. Maybe there's sth wrong in my reasoning? (The professor had different numbers, so maybe there was a typo in his own problem)
I just want to if the way I have solved it makes sense.
Thanx
 
  • #7
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The air resistance would be negative because:

V(spring) = V(gravitational) + friction

Hence, friction would be V(spring) - V(gravitational). Since V(gravitational) > V(spring), friction is negative (ie, friction causes a net gain in energy). Which is the same thing arildno pointed out and I was implying.
 
  • #8
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Hey Parth Dave!
In fact I need to find the energy dissipated by air friction. So what I think is:

Energy = Change in PE + Change in KE
= (m*g*h final - m*g*h initial) + (0.5*m*Vfinal ^2 - 0.5*m*Vinitial^2)
initial mgh = 0 (h = 0) and final 0.5*m*Vfinal = 0 (v=0)
So we now have only: Energy = m*g*h final - 0.5*m*Vinitial^2
0.5*m*Vinitial^2 = energy of spring = 0.5*k*x^2
Finally: Energy = m*g*h final - 0.5*k*x^2

Then for part 2, it seems suitable applying: mgh = 0.5mv^2
 
  • #9
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:smile: The problems are not as difficult as you think. There's no air friction , but we need to find the energy lost by air friction during ascending. Also no wind or any other factor influencing the trajectory of the dart.
 
  • #10
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Well by the conservation of energy:

Ti + Vi = Tf + Vf + Friction (T = kinetic energy, V = potential) hence,
Vi = 0
Tf = 0
Ti = Vf + friction
friction = Ti - Vf

Or from What you did:
Ti + Vi = Tf + Vf + Friction
0 = Change in T + Change in V + friction
-friction = Change in T + Change in V

Both methods are the same, but you just forgot the negative sign.
 
  • #11
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Thanks Parth Dave,
In fact the answer was positive.
But the second Part is weird. As I told, I applied: mgh = 0.5mv^2 and hence I found v. BUt the answer was wrong. What might be the problem?
 

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