A pinball machine has a spring with spring constant of 125N/m. A 65g ball is placed on the spring and the spring is compressed 5cm. How far along the playing surface makes a 15 degrees angle with the horizontal? ignore friction.
How to do this using Energies and Work done? No Kinematics
To determine how far along the playing surface the ball will travel using energies and work done, we need to consider the potential energy stored in the compressed spring and the work performed by gravity and the normal force as the ball moves along the playing surface.
1. Determine initial potential energy:
The potential energy stored in the spring is given by the formula:
Potential energy (PE) = 0.5 * k * x^2
where k is the spring constant and x is the compression distance.
Plugging in the values:
PE = 0.5 * 125 N/m * (0.05 m)^2
PE = 0.15625 J
2. Determine final potential energy:
As the ball reaches the bottom of the incline, all of the potential energy is converted into kinetic energy.
Therefore, the final potential energy is zero. (PE = 0 J)
3. Determine gravitational potential energy:
The ball is lifted to a height along the incline. The change in gravitational potential energy is given by:
∆PE = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
The height can be calculated using the given angle (θ) and the compression distance (x):
h = x * sin(θ)
Plugging in the values:
∆PE = 0.065 kg * 9.8 m/s^2 * 0.05 m * sin(15°)
∆PE ≈ 0.0319 J
4. Calculate the distance along the playing surface:
The work done by gravity and the normal force is equal to the change in potential energy (∆PE):
Work done = ∆PE
Since there is no friction, the work done by the normal force is zero.
Therefore, the work done by gravity is ∆PE.
The work done by gravity can be calculated using the formula:
Work done = m * g * d
where d is the distance along the playing surface.
Plugging in the values and rearranging the equation:
∆PE = 0.065 kg * 9.8 m/s^2 * d
Solving for d:
d = ∆PE / (m * g)
d = 0.0319 J / (0.065 kg * 9.8 m/s^2)
d ≈ 0.049 m
Therefore, the ball will travel approximately 0.049 meters along the playing surface.