A 35.1 g pinball starts at the top of a frictionless pinball table inclined at 46.0°. If the ball is released and the spring at the bottom is compressed 0.0600 m and has a spring constant of k = 27.7 N/m, how far back up the pinball table will the pinball travel before rolling back down again?
To find how far back up the pinball table the pinball will travel before rolling back down again, we can use the conservation of mechanical energy.
The initial potential energy of the pinball can be calculated using the formula:
PE_initial = m * g * h
where m is the mass of the pinball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the table.
The final potential energy of the pinball when it reaches its highest point after rolling back up can be calculated using the formula:
PE_final = 1/2 * k * x^2
where k is the spring constant and x is the displacement of the spring.
Since there is no friction in the system, the mechanical energy is conserved, therefore:
PE_initial = PE_final
m * g * h = 1/2 * k * x^2
Now we can rearrange the equation to solve for x:
x = sqrt((2 * m * g * h) / k)
Substituting the given values:
m = 35.1 g = 0.0351 kg
g = 9.8 m/s^2
h = 0.0600 m
k = 27.7 N/m
x = sqrt((2 * 0.0351 kg * 9.8 m/s^2 * 0.0600 m) / 27.7 N/m)
Calculating the value of x:
x ≈ 0.136 m
Therefore, the pinball will travel approximately 0.136 meters (or 13.6 cm) back up the pinball table before rolling back down again.