To solve this problem, we need to analyze the motion of the particle under the influence of the given force. The force acting on the particle is described by the equation \( F = -\frac{k}{2} x^2 \), where \( k = 0.01 \, \text{Nm}^2 \). This force is a function of the position \( x \), indicating that it is a non-linear force. Let's break down the steps to find the velocity when the particle reaches \( x = 0.5 \, \text{m} \) and the time it takes to reach \( x = 0.25 \, \text{m} \).
Finding the Velocity at x = 0.5 m
First, we can use the work-energy principle, which states that the work done on the particle is equal to the change in its kinetic energy. The work done by the force as the particle moves from \( x = 1.0 \, \text{m} \) to \( x = 0.5 \, \text{m} \) can be calculated using the integral of the force over the distance:
Work done, \( W = \int_{x_1}^{x_2} F \, dx = \int_{1.0}^{0.5} -\frac{0.01}{2} x^2 \, dx \)
Calculating this integral:
- First, find the antiderivative: \( -\frac{0.01}{2} \cdot \frac{x^3}{3} = -\frac{0.01}{6} x^3 \)
- Now evaluate from \( 1.0 \) to \( 0.5 \):
W = \left[-\frac{0.01}{6} (0.5)^3\right] - \left[-\frac{0.01}{6} (1.0)^3\right>
Calculating the values:
- At \( x = 0.5 \): \( -\frac{0.01}{6} \cdot 0.125 = -\frac{0.00125}{6} = -0.0002083 \, \text{J} \)
- At \( x = 1.0 \): \( -\frac{0.01}{6} \cdot 1 = -\frac{0.01}{6} = -0.0016667 \, \text{J} \)
Thus, the work done is:
W = -0.0002083 + 0.0016667 = 0.0014584 \, \text{J}
Now, applying the work-energy theorem:
W = \Delta KE = KE_f - KE_i
Since the initial velocity \( v_i = 0 \), the initial kinetic energy \( KE_i = 0 \). Therefore:
0.0014584 = \frac{1}{2} m v_f^2
Substituting \( m = 0.01 \, \text{kg} \):
0.0014584 = \frac{1}{2} (0.01) v_f^2
Solving for \( v_f^2 \):
v_f^2 = \frac{0.0014584 \times 2}{0.01} = 0.29168
Taking the square root gives:
v_f = \sqrt{0.29168} \approx 0.54 \, \text{m/s}
Determining the Time to Reach x = 0.25 m
To find the time it takes to reach \( x = 0.25 \, \text{m} \), we can use the equations of motion. However, since the force is position-dependent, we will use the relationship between acceleration, force, and mass:
From Newton's second law, \( F = ma \), we have:
a = \( \frac{F}{m} = -\frac{k}{2m} x^2 \)
Substituting the values:
a = \( -\frac{0.01}{2 \times 0.01} x^2 = -\frac{1}{2} x^2 \)
This indicates that the acceleration is proportional to the square of the position. To find the time, we can use the kinematic equation that relates position, initial velocity, acceleration, and time:
x = x_0 + v_0 t + \frac{1}{2} a t^2
Substituting \( x_0 = 1.0 \, \text{m} \), \( v_0 = 0 \), and \( a = -\frac{1}{2} x^2 \):
0.25 = 1.0 + 0 + \frac{1}{2} \left(-\frac{1}{2} x^2\right) t^2
Rearranging gives us a more complex equation to solve, which may require numerical methods or further simplifications. However, for small displacements, we can approximate the time using the average velocity:
Average velocity \( v_{avg} = \frac{v_i + v_f}{2} \approx \frac{0 + 0.54}{2} = 0.27 \, \text{m/s} \)
Using this average velocity to estimate the time taken to travel from \( x = 1.0 \, \text{m} \) to \( x = 0.25 \, \text{m} \):
Distance = \( 1.0 - 0.25 = 0.75 \, \text{m} \)
Time = \( \frac{\text{Distance}}{v_{avg}} = \frac{0.75}{0.27} \approx 2.78 \, \text{s} \)
In summary, the velocity of the particle when it reaches \( x = 0.5 \, \text{m} \) is approximately \( 0.54 \, \text{m/s} \), and the estimated time to reach \( x = 0.25 \, \text{m} \) is about \( 2.78 \, \text{s} \).
