A 0.260 kg block on a vertical spring with a spring constant of 5.00x10^3 N/m is pushed downward, compressing the spring 0.100 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
It travels up until the gravitational energy gain, m g h, equals the potential energy of the compressed spring,
(1/2) k X^2.
Solve for h
sdfsf
To find how high the block rises above the point of release, we need to apply the conservation of mechanical energy.
First, let's find the potential energy stored in the spring when it is compressed. The potential energy in a spring is given by the equation:
PE_spring = (1/2) * k * x^2
where k is the spring constant and x is the compression or extension of the spring.
Substituting the given values:
PE_spring = (1/2) * (5.00x10^3 N/m) * (0.100 m)^2
PE_spring = 25 J
Next, let's find the initial potential energy of the block when it is released. Since the block initially leaves the spring and travels upward, the initial potential energy is equal to the potential energy stored in the spring:
PE_initial = 25 J
As the block rises above the point of release, its potential energy is converted into gravitational potential energy. We can find the maximum height the block reaches by equating the initial potential energy to the gravitational potential energy at the maximum height:
PE_initial = m * g * h_max
where m is the mass of the block, g is the acceleration due to gravity, and h_max is the maximum height reached by the block.
Solving for h_max:
h_max = PE_initial / (m * g)
Substituting the given values:
h_max = 25 J / (0.260 kg * 9.8 m/s^2)
h_max ≈ 9.86 m
Therefore, the block rises approximately 9.86 meters above the point of release.