A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of 40 N/m. The block is initially displaced 4.0 cm from the equilibrium point and then released to set up a simple harmonic motion. What is the speed of the block when it passes through the equilibrium point?
0.57 m/s
I have the answer, but I need help with the steps.
To find the speed of the block when it passes through the equilibrium point, you can use the principle of conservation of mechanical energy.
The mechanical energy of the block-spring system is given by:
E = 1/2 kx² + 1/2 mv²
where E is the total mechanical energy, k is the spring constant, x is the displacement from the equilibrium point, m is the mass of the block, and v is the speed of the block.
At the equilibrium point, the displacement x is zero, so the mechanical energy only consists of the kinetic energy term:
E = 1/2 mv²
To find the speed, we can equate the initial mechanical energy of the system to the mechanical energy at the equilibrium point:
1/2 kx² + 1/2 mv² = 1/2 mv_eq²
where v_eq is the speed at the equilibrium point.
Since the block is initially displaced 4.0 cm from the equilibrium point (0.04 m), we can substitute the values in:
1/2 (40 N/m)(0.04 m)² + 1/2 (0.20 kg)v² = 1/2 (0.20 kg)v_eq²
Simplifying the equation:
0.08 N m + 0.1 kg v² = 0.1 kg v_eq²
Now, we can solve for v_eq:
0.1 kg v_eq² = 0.08 N m + 0.1 kg v²
Dividing both sides of the equation by 0.1 kg:
v_eq² = 0.8 N/m + v²
Substituting the given values:
v_eq² = 0.8 N/m + (0.20 kg)v²
Since the block is at the equilibrium point, its displacement is zero, so the speed at the equilibrium point can be found by solving for v_eq:
v_eq = sqrt(0.8 N/m)
Calculating the square root:
v_eq ≈ 0.89 m/s
Therefore, the speed of the block when it passes through the equilibrium point is approximately 0.89 m/s.