A 1.00 kg glider attached to a spring with a spring constant of 40 N/m oscillates on a horizontal, frictionless air track. At t = 0 the glider is released from rest at x = -17 cm. Find the maximum value of the velocity in m/s.
To find the maximum value of the velocity of the glider, we can use the principle of conservation of mechanical energy. The total mechanical energy of the glider-spring system remains constant as the glider oscillates back and forth.
At any given position, the total mechanical energy is the sum of potential energy (PE) and kinetic energy (KE). The potential energy of the spring, given by Hooke's Law, is PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.
At the rest position (where the glider is released), the potential energy is maximum, and the kinetic energy is zero. As the glider moves away from the equilibrium position and the potential energy decreases, the kinetic energy increases. At the maximum displacement (amplitude), the potential energy is zero, and all the energy is in the form of kinetic energy. At this point, the velocity of the glider is maximum.
We can calculate the maximum displacement by converting the given position from centimeters to meters:
x = -17 cm = -0.17 m
We can now calculate the maximum potential energy of the spring:
PE = (1/2)kx^2
PE = (1/2)(40 N/m)(-0.17 m)^2
PE = 0.578 N·m
Since potential energy is zero at maximum displacement, the total mechanical energy at the maximum displacement is equal to the kinetic energy:
KE = PE = 0.578 N·m
Now we can use the equation for kinetic energy to find the maximum velocity:
KE = (1/2)mv^2
0.578 N·m = (1/2)(1.00 kg)v^2
Solving for v:
v^2 = (2 * 0.578 N·m) / 1.00 kg
v^2 = 1.156 N·m / 1.00 kg
v^2 = 1.156 m^2/s^2
v = √(1.156)
v ≈ 1.076 m/s
Therefore, the maximum value of the velocity of the glider is approximately 1.076 m/s.