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21 June, 10:01

If the spring of a Jack-in-the-Box is compressed a distance of 8 cm from its relaxed length and then released what is the speed of the toy head when the spring returns to its natural length? Assume the mass of the toy head is 50 g the spring constant is 80 N/m, The toy head news only in the vertical direction. Also disregard the mass of the spring. (Hint: remember that there are two forms of potential energy in the problem.)

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  1. 21 June, 13:10
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    The potential energy stored in the compressed spring is converted to the gravitational potential energy of the toy when it moves in the vertical direction upon release, and the kinetic energy of the toy. The potential energy of the compressed spring is (1/2) (k) (x) ^2 = (1/2) (80 N/m) (0.08 m) ^2 = 0.256 J. The gravitational potential energy of the toy when it is at the spring's relaxed length is mgh = (0.05 kg) x (9.81) x (0.08m) = 0.04 J. Hence the kinetic energy of toy at relaxed length is (3.2 - 0.04) J = (0.256 - 0.04) J = 0.216 J. Hence the velocity of the toy head is 0.216 J = (0.5) (0.05 kg) (v^2), v = sqrt (0.216J/0.025 kg) = 2.94 m/s.
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