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techninja
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A child's pogo stick (see figure) stores energy in a spring of spring constant k = 40000 N/m. At position A, x1 = -0.150 m, the spring compression is a maximum and the child is momentarily at rest. At position B, the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. Assuming that the combined mass of child and pogo stick is 26.0 kg, answer the following:
(a) Calculate the total energy of the system if both potential energies are zero at x = 0
I did this through:
[tex]
ME_i = U_s + U_g = \frac{1}{2}kx^2+mgh = 411.78 J
[/tex]
(b) Determine x2
I did this through:
[tex]
ME_i = ME_f = U_g
ME_i = mgh[/tex]
[tex]
x_2 = \frac{ME_i}{mg} = 1.62 m
[/tex]
(c) Calculate the speed of the child at x = 0
I did this through:
[tex]
ME_i = \frac{1}{2}mv^2
v = \sqrt{\frac{2ME_i}{m}} = 5.63 m/s
[/tex]
(d) Determine the value of x for which the kinetic energy of the system is a maximum.
Hint: The kinetic energy (and velocity) is a maximum when the acceleration is zero at that point.
This one, I tried very hard... but I can't do it.
I get:
[tex]
v = \sqrt{\frac{2(1378-mgh)}{m}}
a = \frac{1}{m\sqrt{\frac{2h}{m}}}
[/tex]
Tex hates me. It should look something like v = sqrt([2(1378-mgh)]/m), so a = 1/(m*sqrt(2h/m))
Which doesn't seem to work out nicely.
(e) Obtain the child's maximum upward speed
I think I could do this with information from D, but ... I'm not sure.
Any help is muchly appreciated!
A child's pogo stick (see figure) stores energy in a spring of spring constant k = 40000 N/m. At position A, x1 = -0.150 m, the spring compression is a maximum and the child is momentarily at rest. At position B, the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. Assuming that the combined mass of child and pogo stick is 26.0 kg, answer the following:
(a) Calculate the total energy of the system if both potential energies are zero at x = 0
I did this through:
[tex]
ME_i = U_s + U_g = \frac{1}{2}kx^2+mgh = 411.78 J
[/tex]
(b) Determine x2
I did this through:
[tex]
ME_i = ME_f = U_g
ME_i = mgh[/tex]
[tex]
x_2 = \frac{ME_i}{mg} = 1.62 m
[/tex]
(c) Calculate the speed of the child at x = 0
I did this through:
[tex]
ME_i = \frac{1}{2}mv^2
v = \sqrt{\frac{2ME_i}{m}} = 5.63 m/s
[/tex]
(d) Determine the value of x for which the kinetic energy of the system is a maximum.
Hint: The kinetic energy (and velocity) is a maximum when the acceleration is zero at that point.
This one, I tried very hard... but I can't do it.
I get:
[tex]
v = \sqrt{\frac{2(1378-mgh)}{m}}
a = \frac{1}{m\sqrt{\frac{2h}{m}}}
[/tex]
Tex hates me. It should look something like v = sqrt([2(1378-mgh)]/m), so a = 1/(m*sqrt(2h/m))
Which doesn't seem to work out nicely.
(e) Obtain the child's maximum upward speed
I think I could do this with information from D, but ... I'm not sure.
Any help is muchly appreciated!
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