A 4.9 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.0 cm if the marble is to just reach a target 29 m above the marble's position on the compressed spring.
(a) What is the change in the gravitational potential energy of the marble-Earth system during the 29 m ascent?
1 J
(b) What is the change in the elastic potential energy of the spring during its launch of the marble?
2 J
(c) What is the spring constant of the spring?
3 N/m
all of your answers are wrong.
To find the answers to these questions, we can use the principles of potential energy and Hooke's Law.
(a) The change in gravitational potential energy (ΔPEg) of the marble-Earth system during the 29 m ascent can be calculated using the formula:
ΔPEg = mgh
where ΔPEg is the change in gravitational potential energy, m is the mass of the marble, g is the acceleration due to gravity, and h is the height change.
Given:
m = 4.9 g = 0.0049 kg (Convert grams to kilograms)
g = 9.8 m/s^2 (Approximate value for acceleration due to gravity)
h = 29 m
Using the formula, we can find:
ΔPEg = (0.0049 kg) * (9.8 m/s^2) * (29 m)
= 1.3666 J (Approximately 1 J)
Therefore, the change in gravitational potential energy of the marble-Earth system during the 29 m ascent is approximately 1 J.
(b) The change in elastic potential energy (ΔPEs) of the spring during its launch of the marble can be calculated using the formula:
ΔPEs = (1/2)kΔx^2
where ΔPEs is the change in elastic potential energy, k is the spring constant, and Δx is the displacement of the spring.
Given:
Δx = 8.0 cm = 0.08 m (Convert centimeters to meters)
Since we don't have the value of ΔPEs, we need to find the value of the spring constant k.
(c) The spring constant (k) can be found using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement:
F = kΔx
Given:
F = mg (Since the spring force is balanced by the weight of the marble during launch)
Using the known values from earlier:
m = 0.0049 kg
Δx = 0.08 m
Substituting these values into Hooke's Law, we have:
mg = kΔx
k = (mg) / Δx
= (0.0049 kg) * (9.8 m/s^2) / (0.08 m)
= 0.6 N/m (Approximately 3 N/m)
Therefore, the spring constant of the spring is approximately 3 N/m.
To summarize:
(a) The change in gravitational potential energy of the marble-Earth system during the 29 m ascent is approximately 1 J.
(b) The change in elastic potential energy of the spring during its launch of the marble is not given in the provided information.
(c) The spring constant of the spring is approximately 3 N/m.