Ideal Toy

Ideal Toy

Big Boy Toys Can Bring a Smile

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Spring Toy mass 0.65g pressed down onto table till it sticks: compression of spring = 1.2cm. Three Questions:?

When the vacuum seal that the toy makes with the table fails, it rises 34 cm vertically.

Given this and the other information in the question title:

A) Assuming the spring is ideal and that air resistance and other losses are negligible, calculate the spring constant of the spring.

B) What is the magnitude of the force that has to be applied to compress the spring sufficiently for the toy to be pressed to the table?

C) Calculate the speed with which the toy leaves the table

Any help gratefully recieved!

max P.E. of toy = mgh {where m = 0.65e-3, g = 9.81, h = 0.34}
max P.E. = (0.65e-3)(9.81)(0.34)
max P.E. = 2.17e-3 J

Spring P.E. = 2.17e-3 J
Spring P.E. = 1/2 kx²
2.17e-3 = (0.5)(k)(1.2e-2)²
2.17e-3 = 0.72e-4 (k)
k = 2.17e-3/0.72e-4 = 3.01e1 = 30.1 N/m ANS-A
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Work to compress toy spring = 2.17e-3 J
Work = 1/2 k(x²) = 2.17e-3
(0.5)(30.1)(x²) = 2.17e-3
x² = 2.17e-3/(0.5)(30.1)
x² = 0.144e-3 = 1.44e-4
x = 1.2e-2 m

F = kx
F = (30.1)(1.2e-2) = 36.12e-2 = 0.36 N ANS-B
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max P.E.{at max height} = max K.E. {at table top}
2.17e-3 = 1/2mV² {where m = 0.65e-3}
2.17e-3 = (0.5)(0.65e-3)V²
cancel e-3
2.17 = 0.325V²
V² = 2.17/0.325
V² = 6.68
V = 2.58 m/s ANS-C

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