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A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

0.003 kg and the spring constant is 220 N/m. What is the maximum height the marble can reach (a) when

compressed 2 cm? (b) when compressed 4 cm?​

User Macetw
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2 Answers

1 vote

Final answer:

To find the maximum height the marble can reach when compressed, we can equate the potential energy stored in the spring to the gravitational potential energy at the maximum height. For a compression of 2 cm, the marble can reach a maximum height of approximately 0.747 m, and for a compression of 4 cm, the marble can reach a maximum height of approximately 5.8 m.

Step-by-step explanation:

To find the maximum height the marble can reach, we can use the concept of mechanical energy conservation. When the spring is compressed, it contains potential energy which is transferred to the marble when it is launched. At the maximum height, all of the potential energy is converted to gravitational potential energy, given by the equation mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

(a) When the spring is compressed 2 cm (0.02 m), the potential energy is given by 1/2 kx^2, where k is the spring constant and x is the compression. So, the potential energy is 1/2(220 N/m)(0.02 m)^2 = 0.022 J. Equating this to mgh, we have 0.022 J = (0.003 kg)(9.8 m/s^2)h. Solving for h gives us h = 0.022 J / (0.003 kg)(9.8 m/s^2) ≈ 0.747 m.

(b) When the spring is compressed 4 cm (0.04 m), the potential energy is given by 1/2(220 N/m)(0.04 m)^2 = 0.176 J. Equating this to mgh, we have 0.176 J = (0.003 kg)(9.8 m/s^2)h. Solving for h gives us h = 0.176 J / (0.003 kg)(9.8 m/s^2) ≈ 5.8 m.

User Fracca
by
5.1k points
5 votes

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Step-by-step explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have


(1)/(2)kx^2 = mg(H + x)


0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have


H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have


(1)/(2)kx^2 = mg(H + x)


0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have


H = 5.94 m

User SeaBean
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