The potential energy of a 2.0-kg particle moving along the x axis is given by
U(x) = 5.0 x^2.
When the particle is at x = 1.0 m it is traveling in the positive x direction with a speed of 1.5 m/s. It next stops momentarily to turn around at x =
To find the point where the particle stops momentarily to turn around, we need to find the point on the x-axis where the particle's kinetic energy is zero. At this point, all of its energy is in the form of potential energy.
Given:
Particle mass (m) = 2.0 kg
Potential energy function (U(x)) = 5.0x^2
Initial position (x) = 1.0 m
Initial velocity (v) = 1.5 m/s
We can start by finding the total mechanical energy of the particle, which is the sum of its kinetic and potential energy:
E = K + U
Where:
E = Total mechanical energy
K = Kinetic energy
U = Potential energy
Given that the particle is moving, its kinetic energy can be calculated using the formula:
K = (1/2)mv^2
Substituting the known values:
K = (1/2)(2.0 kg)(1.5 m/s)^2
K = (1/2)(2.0 kg)(2.25 m^2/s^2)
K = 2.25 J
Now, we can find the total mechanical energy by substituting the values of K and U into the equation:
E = K + U
E = 2.25 J + 5.0x^2
Since at the point where the particle stops momentarily, the kinetic energy is zero, the total mechanical energy is equal to the potential energy:
E = U
2.25 J + 5.0x^2 = 5.0x^2
Simplifying the equation:
2.25 J = 0
It is important to note that there is no real number solution to this equation. This means that the particle does not stop momentarily to turn around at any point on the x-axis.
To find the point at which the particle stops momentarily to turn around, we need to look for the point at which the particle's kinetic energy is zero.
The kinetic energy (K) of the particle is given by the equation:
K = (1/2)mv^2
where m is the mass of the particle and v is its velocity.
Given that the particle is moving along the x axis, we can find its velocity by taking the derivative of the potential energy equation with respect to time:
v = dx/dt
Now, let's find the force acting on the particle using the second law of motion, which states that the force (F) is equal to the negative gradient (slope) of the potential energy function:
F = -dU/dx
Taking the derivative of the potential energy function U(x) = 5.0x^2, we get:
dU/dx = 10.0x
Let's equate this to the force acting on the particle:
F = -10.0x
Since the force is related to the acceleration (a) of the particle through Newton's second law (F = ma), the equation becomes:
ma = -10.0x
Now, let's substitute the expression for velocity (v = dx/dt) and the expression for acceleration (a = dv/dt) into the equation:
m(dv/dt) = -10.0x
Rearranging the equation, we get:
mdv = -10.0xdx
Integrating both sides with respect to their respective variables, we obtain:
∫mdv = -10.0∫xdx
Integrating both sides, we get:
mv = -5.0x^2 + C
Where C is the constant of integration.
Given that at x = 1.0 m, the mass (m) of the particle is 2.0 kg and the velocity (v) is 1.5 m/s, we can substitute these values into the equation:
(2.0 kg)(1.5 m/s) = -5.0(1.0 m)^2 + C
Simplifying the equation, we find the value of C:
C = 4.5 N·m^2/s^2 + 10 J
Now, let's set the kinetic energy (K) equal to zero to find the point at which the particle stops momentarily:
K = (1/2)mv^2 = 0
(1/2)(2.0 kg)(v^2) = 0
v^2 = 0
v = 0
Since the velocity is zero, we can substitute this value into the equation for velocity (mv = -5.0x^2 + C) and solve for the value of x:
(2.0 kg)(0) = -5.0(x)^2 + (4.5 N·m^2/s^2 + 10 J)
0 = -5.0(x)^2 + 4.5 N·m^2/s^2 + 10 J
Rearranging the equation, we have:
5.0(x)^2 = 4.5 N·m^2/s^2 + 10 J
Dividing both sides by 5.0, we get:
(x)^2 = (4.5 N·m^2/s^2 + 10 J) / 5.0
Simplifying the equation, we find:
(x)^2 = 0.9 N·m^2/s^2 + 2 J
Taking the square root of both sides, we find:
x = sqrt(0.9 N·m^2/s^2 + 2 J)
Therefore, the particle stops momentarily to turn around at x = sqrt(0.9 N·m^2/s^2 + 2 J).