potential energy well for a 2.0 kg particle (the function U(x) has the form bx2 and the vertical axis scale is set by us = 2.0 J).
(a) If the particle passes through the equilibrium position with a velocity of 85 cm/s, will it be turned back before it reaches x = 15 cm?
(b) If yes, at what position, and if no, what is the speed of the particle at x = 15cm?
To determine if the particle will be turned back before it reaches x = 15 cm, we need to find the turning points of the potential energy function and compare them to the position x = 15 cm.
(a) Find the turning points:
The turning points occur where the derivative of the potential energy function U(x) with respect to x is zero:
U'(x) = 2bx = 0
Setting U'(x) = 0, we find x = 0. Therefore, the equilibrium position is x = 0.
(b) Determine if the particle will be turned back:
To determine if the particle will be turned back, we need to calculate the particle's total mechanical energy at the equilibrium position and at x = 15 cm. If the total mechanical energy is greater at x = 15 cm, the particle will not be turned back.
The total mechanical energy (E) is given by the sum of kinetic energy (K) and potential energy (U):
E = K + U
At the equilibrium position x = 0, the potential energy is the maximum, and the kinetic energy is zero (since the particle is momentarily at rest). Therefore, the total mechanical energy at x = 0 is:
E(0) = K(0) + U(0) = 0 + U(0) = 2 J
At x = 15 cm, we need to calculate the kinetic energy. Since we know the mass of the particle (2.0 kg) and its velocity (85 cm/s), we can calculate the kinetic energy as follows:
K(15 cm) = (1/2)mv^2 = (1/2)(2.0 kg)(85 cm/s)^2 = 1442.5 J
Now, let's compare the total mechanical energies at x = 0 and x = 15 cm:
E(15 cm) = K(15 cm) + U(15 cm) = 1442.5 J + b(15 cm)^2
If E(15 cm) is greater than E(0), the particle will not be turned back.
Setting E(15 cm) > E(0), we have:
1442.5 J + b(15 cm)^2 > 2 J
Simplifying the inequality, we get:
b(15 cm)^2 > -1440.5 J
Since b and the position (x) are both positive, the inequality is always true. Therefore, the particle will not be turned back before it reaches x = 15 cm.
(b) The speed of the particle at x = 15 cm:
If the particle is not turned back, it will continue moving past x = 15 cm. To find the particle's speed at x = 15 cm, we can calculate the magnitude of its velocity.
Using the total mechanical energy at x = 15 cm, we have:
E(15 cm) = K(15 cm) + U(15 cm)
Rearranging the equation, we can solve for the kinetic energy at x = 15 cm:
K(15 cm) = E(15 cm) - U(15 cm) = 1442.5 J + b(15 cm)^2 - (2 J)
Substituting the values, we have:
K(15 cm) = 1440.5 J + b(15 cm)^2
To find the speed, we can use the relationship between kinetic energy and speed:
K(15 cm) = (1/2)mv^2
Rearranging the equation and solving for v, we have:
v = sqrt(2K(15 cm)/m)
Substituting the values, we get:
v = sqrt(2(1440.5 J + b(15 cm)^2)/(2.0 kg))
To answer these questions, we need to determine the equation for the potential energy function U(x) and then use the conservation of mechanical energy to find the answers.
(a) The potential energy function U(x) has the form bx^2, and the vertical axis scale is set by U_s = 2.0 J. This means that when x = 1 meter (100 cm), U(x) = 2.0 J.
To find the constant b, we can use the given information about the equilibrium position and the known velocity. When the particle is at the equilibrium position, the potential energy is at its minimum and the kinetic energy is at its maximum. Therefore, the total mechanical energy at the equilibrium position is given by:
E_total = U(x_eq) + K_eq
Since the particle is at rest at the equilibrium position, the kinetic energy K_eq is zero. Thus, E_total = U(x_eq).
Since U(x) = bx^2, we have:
E_total = U(x_eq) = b(x_eq)^2
But we know that U_s = 2.0 J corresponds to x = 1 meter (100 cm). Therefore, we can write:
E_total = b(x_eq)^2 = U_s = 2.0 J
Now, we can solve for b:
b(x_eq)^2 = 2.0 J
b(100 cm)^2 = 2.0 J
b = 2.0 J / (100 cm)^2 = 2.0 J / 10000 cm^2 = 2.0 × 10^-4 J / cm^2
Now, we have the potential energy function U(x) for the given system.
(b) To determine whether the particle will be turned back before reaching x = 15 cm, we need to compare the total mechanical energy E_total at the given position with the potential energy U(x).
First, let's find the total mechanical energy E_total at x = 15 cm:
E_total = U(15 cm) + K(15 cm)
Since the particle is not at rest at x = 15 cm, we need to know its velocity to calculate the kinetic energy K(15 cm). However, the velocity is not given in the question.
If the particle's velocity is given, we can calculate the kinetic energy using K = (1/2)mv^2, where m is the mass of the particle and v is its velocity. Then, we can compare E_total with U(15 cm) to determine whether the particle will be turned back or not.
If the particle's velocity is not given, we cannot determine whether the particle will be turned back before reaching x = 15 cm or not.