A student finds that a stretched rubber band stores 3.5 J of energy. She uses it to propel an object of mass 0.025 kg vertically upwards. What will be the initial speed of the object? How high can it travel before coming momentarily to rest?
Use the principles and equations I have shown you in my other answers. We will be glad to critique your work.
To find the initial speed of the object propelled by the rubber band, we need to use the principle of conservation of energy. The stored energy in the rubber band is converted into the kinetic energy (initial speed) of the object.
The formula for kinetic energy is given by:
KE = 1/2 * m * v^2
Where KE is the kinetic energy, m is the mass of the object, and v is the velocity (initial speed).
Given:
Stored energy in the rubber band (E) = 3.5 J
Mass of the object (m) = 0.025 kg
To find the initial speed (v):
KE = 1/2 * m * v^2
3.5 J = 1/2 * 0.025 kg * v^2
v^2 = (3.5 J) / [(1/2) * 0.025 kg]
v^2 = (3.5 J) / (0.0125 kg)
v^2 = 280 m^2/s^2
v = √(280 m^2/s^2)
v ≈ 16.733 m/s
Therefore, the initial speed of the object propelled by the rubber band is approximately 16.733 m/s.
Now, let's find how high the object can travel before coming momentarily to rest. We can use the principle of conservation of mechanical energy, where the potential energy at the highest point is equal to the initial kinetic energy.
The potential energy (PE) at a certain height is given by:
PE = m * g * h
Where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
At the highest point, the potential energy equals the initial kinetic energy:
PE = KE
m * g * h = 1/2 * m * v^2
Substituting the given values:
0.025 kg * 9.8 m/s^2 * h = 1/2 * 0.025 kg * (16.733 m/s)^2
0.245 kg * h = 1/2 * 0.025 kg * 280.546289 m^2/s^2
0.245 kg * h = 3.507 m^2/s^2
h ≈ (3.507 m^2/s^2) / (0.245 kg)
h ≈ 14.3 m
Therefore, the object can travel approximately 14.3 meters high before coming momentarily to rest.