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?
Well, isn't this a spring-loaded glider on an air track just spring-tacular? Let's unravel this conundrum.
We know that the glider is released from rest, so at that point, its potential energy, stored in the compressed spring, is converted into kinetic energy. Now, when the speed of the glider reaches 0.20 m/s, all of that initial potential energy has transformed into kinetic energy.
Since no external forces are acting upon the glider (besides the spring force), the total mechanical energy of the system (glider + spring) must be conserved. So, we have the equation:
1/2mv^2 + 1/2kx^2 = 1/2kA^2
Where m is the mass of the glider (0.200 kg), v is the final velocity (0.20 m/s), k is the spring constant (5.00 N/m), x is the displacement from the equilibrium position, and A is the initial displacement (0.100 m).
Now let's solve for x:
1/2 * 0.200 kg * (0.20 m/s)^2 + 1/2 * 5.00 N/m * x^2 = 1/2 * 5.00 N/m * (0.100 m)^2
Now, let's simplify this equation and solve for x:
0.020 J + 2.50 N/m * x^2 = 0.025 J
2.50 N/m * x^2 = 0.025 J - 0.020 J
2.50 N/m * x^2 = 0.005 J
x^2 = 0.005 J / (2.50 N/m)
x^2 = 0.002 m
x = sqrt(0.002 m)
x = 0.045 m
So, the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s is approximately 0.045 m. Voila!
To find the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s, we can use the principle of conservation of energy.
The total mechanical energy of the system (spring and glider) is conserved, which means that it remains constant throughout the motion.
The total mechanical energy is given by the sum of the potential energy stored in the spring and the kinetic energy of the glider:
E_total = E_potential + E_kinetic
The potential energy stored in the spring is given by:
E_potential = (1/2)kx^2
where k is the force constant and x is the displacement.
The kinetic energy of the glider is given by:
E_kinetic = (1/2)mv^2
where m is the mass of the glider and v is its velocity.
Since the initial position is at the equilibrium position, the potential energy is 0. Therefore, at any other position, the total mechanical energy is equal to the kinetic energy.
E_total = E_kinetic
(1/2)kx^2 = (1/2)mv^2
Substituting the given values:
(1/2)(5.00 N/m)(x)^2 = (1/2)(0.200 kg)(0.20 m/s)^2
Simplifying:
(2.50 N/m)(x)^2 = 0.02 kg m^2/s^2
x^2 = 0.02 kg m^2/s^2 / 2.5 N/m
x^2 = 0.008 m^2
Taking the square root of both sides:
x = √(0.008 m^2)
x = 0.089 m
Therefore, the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s is 0.089 m.
To find the displacement of the glider from its equilibrium position when its speed is 0.20 m/s, we need to use the concepts of energy conservation and Hooke's Law.
1. Energy conservation: Initially, the glider has potential energy stored in the spring when it is stretched. As it moves, this potential energy is converted into kinetic energy. At any point, the total mechanical energy (potential energy + kinetic energy) remains constant.
2. Hooke's Law: The force exerted by the spring is proportional to the displacement from its equilibrium position. This relationship is given by Hooke's Law as F = -k * x, where F is the force, k is the force constant, and x is the displacement.
Let's break down the problem and find the displacement x.
Given:
Mass of the glider, m = 0.200 kg
Force constant of the spring, k = 5.00 N/m
Initial displacement of the spring, x_initial = 0.100 m
Final speed of the glider, v_final = 0.20 m/s
Step 1: Find the initial potential energy of the spring.
The initial potential energy of the spring is given by the formula:
Potential energy_initial = (1/2) * k * x_initial²
Substituting the given values:
Potential energy_initial = (1/2) * 5.00 N/m * (0.100 m)²
Step 2: Find the final kinetic energy of the glider.
The final kinetic energy of the glider is given by the formula:
Kinetic energy_final = (1/2) * m * v_final²
Substituting the given values:
Kinetic energy_final = (1/2) * 0.200 kg * (0.20 m/s)²
Step 3: Set the initial potential energy equal to the final kinetic energy.
Potential energy_initial = Kinetic energy_final
(1/2) * 5.00 N/m * (0.100 m)² = (1/2) * 0.200 kg * (0.20 m/s)²
Step 4: Solve for the displacement.
Simplify the equation and solve for x:
x² = (0.200 kg * (0.20 m/s)²) / (5.00 N/m)
x² = 0.016 m²
x = ± 0.04 m
Therefore, the displacement x of the glider from its equilibrium position when its speed is 0.20 m/s is either +0.04 m or -0.04 m.
F on glider from spring = -kx
beginning x = .1 meter
potential energy in spring = kd^2/2 where d is extension
= 5 d^2/2 = 2.5 d^2
= 2.5 (.1-x)^2
the potential energy at start = 2.5(.1)^2 = .025 Joules total
the kinetic energy = (1/2)mv^2
= .1 (.2)^2 = .004 Joules
so
.025 = .004 + 2.5 (.1-x)^2
.0084 = (.1-x)^2
.0917 = .1-x
x = .00835 meter
check my algebra