A glider with mass m=0.200kg sits on a frictionless horizontal air track, connected to a spring with force constant k=5.00N/m. The glider is released from rest with the spring stretched 0.100 m.
-What is the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s?
1/2 k x^2 = 1/2 m v^2
so x = sqrt(m v^2/k) = sqrt(.2*.2^2/5)
To find the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s, we can use the concept of energy conservation.
The glider's initial potential energy when the spring is stretched is given by U_initial = (1/2)kx^2, where x is the initial displacement. The glider's initial kinetic energy is zero.
When the glider is released, the potential energy is converted into kinetic energy, given by K = (1/2)mv^2, where m is the mass of the glider and v is its speed. Since the glider is released from rest, its initial kinetic energy is zero.
According to energy conservation, the initial potential energy is equal to the final kinetic energy. Therefore, we can equate the two equations:
(1/2)kx^2 = (1/2)mv^2
To solve for x, let's rearrange the equation:
x^2 = (mv^2) / k
x = sqrt((mv^2) / k)
Plugging in the given values:
m = 0.200 kg
v = 0.20 m/s
k = 5.00 N/m
x = sqrt((0.200 kg * (0.20 m/s)^2) / 5.00 N/m)
x = sqrt((0.200 kg * 0.04 m^2) / 5.00 N/m)
x = sqrt(0.008 kg⋅m^2 / 5.00 N)
Calculating the value:
x ≈ 0.089 m
Therefore, the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s is approximately 0.089 m.