A spring and block is set up on frictionless tabletop. The 30N/m spring is compressed 16 centimeters and released. What is the maximum speed the 2.0 kilogram block will reach?
To find the maximum speed the block will reach, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the spring will be converted into kinetic energy when the spring is released.
First, let's convert the displacement from centimeters to meters:
16 centimeters = 16/100 = 0.16 meters
Next, we can calculate the potential energy stored in the spring when it is compressed:
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement.
Given that k = 30 N/m and x = 0.16 m, we can plug in the values:
PE = (1/2) * 30 N/m * (0.16 m)^2
= 0.24 J
Since energy is conserved, this potential energy will be transferred to kinetic energy when the spring is released. The maximum speed (v) can be calculated using the formula for kinetic energy:
Kinetic Energy (KE) = (1/2) * m * v^2
where m is the mass of the block, and v is the velocity.
Given that m = 2.0 kg, we can rearrange the equation to solve for v:
KE = (1/2) * m * v^2
0.24 J = (1/2) * 2.0 kg * v^2
0.24 J = 1.0 kg * v^2
Dividing both sides of the equation by 1.0 kg:
v^2 = 0.24 J / 1.0 kg
v^2 = 0.24 m^2/s^2
Finally, taking the square root of both sides:
v ≈ √(0.24 m^2/s^2)
v ≈ 0.49 m/s
Therefore, the maximum speed the 2.0 kilogram block will reach is approximately 0.49 m/s.