To determine how far the block moves down the incline from its release point to the stopping point, we can use the concept of conservation of mechanical energy.
The potential energy lost by the block as it moves down the incline is converted into the potential energy stored in the compressed spring.
The potential energy lost by the block is given by:
ΔPE = m * g * h
Where:
m is the mass of the block (12.0 kg)
g is the acceleration due to gravity (approximately (9.8 m/s^2)
h is the vertical distance traveled by the block down the incline
The potential energy stored in the compressed spring is given by:
PE_spring = (1/2) * k * x^2
Where:
k is the spring constant 1.55 × 10^4 N/m
x is the compression of the spring 5.50 cm
Since the block momentarily stops at the stopping point, the potential energy lost by the block is equal to the potential energy stored in the compressed spring:
ΔPE = PE_spring
Therefore, we can equate the two expressions:
m * g * h = (1/2) * k * x^2
Substituting the given values:
m = (12.0 kg)
g = (9.8 m/s^2)
k = (1.55 × 10^4 N/m)
x = (0.055 m)
We can solve this equation for h, which represents the vertical distance traveled by the block down the incline.
h = (1/2) * (k/m) * x^2 / g
Substituting the values:
h = (1/2) * (1.55 × 10^4 N/m) / (12.0 kg) * (0.055 m)^2 / (9.8 m/s^2)
h ≈ (0.1346 m)
Finally, to convert the distance to centimeters, we multiply by 100:
h ≈ (13.46 cm)
Therefore, the block moves approximately 13.46 centimeters down the incline from its release point to the stopping point!