A spring toy is used to fire a 10.6 gram pellet at a target directly above it. T

Question

A spring toy is used to fire a 10.6 gram pellet at a target directly above it. The target is 3.2 m above the point where the pellet was at rest on the compressed spring. When it reaches the target, half of the original energy stored in the spring remains as kinetic energy of the pellet. The spring constant is 22 N/m. Neglect air resistance. I know the gravitational potential energy of the pellet increases 0.33 J during its flight to the target. I need to know how much potential energy was initially stored in the spring. Take the potential energy of the compressed spring to be zero. The answer should be 0.66 J but I don't get this. Thanks so much.

Explanation / Answer

k = 22 N/m m = .0106 kg h = 3.2 m Vo = 0 m/s g = 9.8 m/s^2 -------------- Total energy in the system is conserved: PE(spring) = PE(Pellet) + KE(Pellet) It tells you that half of the PE(spring) becomes KE(pellet): PE(spring) = PE(Pellet) + .5(PE(spring)) Which reduces to .5(PE(spring)) = PE(pellet) PE = mgh, which you found correctly to be 0.33 J .5(PE(spring)) = .33 J PE(spring) = .66 J That is the potential energy in the spring. You can then figure out how far the spring is compressed using PE(spring) = .5kx^2 where k is given to be 22 N/m

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