A 5 Kg marble is fired vertically upward using a spring gun. The spring must be compressed 8.0cm if the marble is to just reach a target 20m above the marble’s position on the compressed spring.
a) What is the change ∆U in the gravitational potential energy of the marble-Earth system during the 20m ascent?
b) What is the change in ∆U in the elastic potential energy of the spring during its launch of marble?
c) What is the spring constant of the spring?
I am sure that you know by now the formulas for gravitational potential energy and the potential energy of a compressed spring.
The two P.E.s are equal in this case.
To find the answers to these questions, we'll need to use the principles of potential energy and Hooke's Law.
a) The change (∆U) in gravitational potential energy of the marble-Earth system during the 20m ascent can be calculated using the formula:
∆U = m * g * h
Where:
∆U is the change in gravitational potential energy
m is the mass of the marble (5 Kg)
g is the acceleration due to gravity (usually taken as 9.8 m/s^2 on Earth)
h is the height change (20 m)
Plugging in the values, we have:
∆U = 5 Kg * 9.8 m/s^2 * 20 m
∆U = 980 Joules
Therefore, the change in gravitational potential energy (∆U) is 980 Joules.
b) The change (∆U) in the elastic potential energy of the spring during its launch can be calculated using the formula:
∆U = 0.5 * k * x^2
Where:
∆U is the change in the elastic potential energy
k is the spring constant (to be determined)
x is the displacement of the spring (8.0 cm = 0.08 m)
Since the marble reaches a target 20 m above its initial position, we know that the potential energy stored in the compressed spring is fully converted into gravitational potential energy. So, the change in elastic potential energy is equal to the change in gravitational potential energy.
∆U = ∆U_gravitational
0.5 * k * 0.08 m^2 = 980 J
Simplifying the equation, we get:
k = (980 J) / (0.5 * 0.08 m^2)
k = 30,625 N/m
Therefore, the spring constant of the spring is 30,625 N/m.
In summary:
a) The change in gravitational potential energy (∆U) during the 20m ascent is 980 Joules.
b) The change in elastic potential energy (∆U) during the launch of the marble is also 980 Joules.
c) The spring constant of the spring is 30,625 N/m.