Question

A ball bearing of mass m=50.0g, is sitting on a vertical spring whose force constant is 120.0N/m. The initial position of the spring is at y=0m. The spring is compressed downward a distance x=0.200m. From the compressed position, how high will the ball bearing rise? How high does the ball bearing rise above the equilibrium position at y=0m?

Answer 1

Answer on Question #60590, Physics / Mechanics | Relativity

A ball bearing of mass $m = 50.0\mathrm{g}$, is sitting on a vertical spring whose force constant is $120.0\mathrm{N/m}$. The initial position of the spring is at $y = 0\mathrm{m}$. The spring is compressed downward a distance $x = 0.200\mathrm{m}$. From the compressed position, how high will the ball bearing rise? How high does the ball bearing rise above the equilibrium position at $y = 0\mathrm{m}$?

Solution:

Assume frictionless system and massless spring

The potential energy of ball is

$$PE = mgh$$

The potential energy of spring is

$$PS = \frac{1}{2}kx^2$$

The kinetic energy of ball is

$$KE = \frac{1}{2}mv^2$$

A) I assume the question is how high the ball will rise if the spring were released after compression of $0.200\mathrm{m}$

$$PS = \frac{1}{2}kx^2$$ $$PS = \frac{1}{2} \cdot 120 \cdot (0.2)^2 = 2.4\ \mathrm{J}$$

What height would give a $50\ \mathrm{g}$ ball a potential energy of $2.4\ \mathrm{J}$?

$$PE = PS$$ $$PE = mgh$$ $$h = \frac{PE}{mg} = \frac{2.4}{0.05 \cdot 9.8} \approx 4.90\ \mathrm{m}$$

The second question asks how far from our origin of uncompressed spring so we need to subtract the distance of spring compression.

$$h_0 = 4.9 - 0.2 = 4.7\ \mathrm{m}$$

Answer: $4.90\ \mathrm{m}; 4.7\ \mathrm{m}$.

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