Question

A pinball machine uses a spring to launch a small steel ball (m =250 g) up the pinball machine so you can play. The incline of most machines is around 15 degrees and this particular machine is 1 m long. If the ball is calibrated to be moving at 2 m/s at the top of the machine after being launched, and the plunder can depress a maximum of 5 cm, calculate the spring constant of this pinball launcher. A. 90.7 N/m B. 2300 N/m C. 501 N/m D. 907 N/m

Answer 1

Answer : (D) 907 N/m

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Solution :

here we have :

m = 250 g = 0.250 kg

θ = 15^o

L = 1 m

v = 2 m/s

x = 5 cm = 0.05 m

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According to the conservation of energy : KE_final = PE_spring - W_gravity

∴ (1/2) m v² = (1/2) k x² - m g h

here, h = L sinθ

∴ (1/2) m v² = (1/2) k x² - m g (L sinθ)

∴ m v² = k x² - 2m g (L sinθ)

∴ m v² + 2m g (L sinθ) = k x²

∴ (0.250 kg)(2 m/s)² + 2(0.250 kg)(9.80 m/s²)(1 m) sin(15) = k (0.05 m)²

∴ (2.2695 Nm) = k (0.05 m)²

∴ k = 907 N/m