A particle of mass m starts from x subscript o = 0 m with V subscript o > 0 m/s. The particle experiences the variable force F subscript x = F subscript o sin (cx) as it moves to the right along the x-axis, where F subscript o and c are constants.
I have answered the first 3 of 4 questions correctly and keep getting a message of format error in the last problem.
(1)The units of F (subscript o) = N
(2)The units of c = m^-1
(3)At what position x (subscript max) does the force first reach a maximum value? pi/2c
(4)What is the particle's velocity as it reaches x (subscript max)? Give your answer in terms of m, v (subscript o), F (subscript o), and c?
To find the particle's velocity as it reaches x (subscript max), we can start by using Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
Given that the force acting on the particle is F subscript x = F subscript o sin (cx), and the mass of the particle is m, we can express the acceleration of the particle as:
a = F subscript x / m
Since the force acting on the particle varies with position, we need to find the maximum value of the force (F subscript max). To determine this, we can take the derivative of the force function with respect to x and set it equal to zero:
d(F subscript x) / dx = 0
Differentiating the given force equation, we have:
d(F subscript o sin (cx)) / dx = 0
Using the chain rule, we can differentiate the sin(cx) term with respect to x:
F subscript o cos(cx) * c = 0
Simplifying, we find:
cos(cx) = 0
This occurs when cx = pi/2, or x (subscript max) = pi/2c.
Now, we can determine the particle's velocity at x (subscript max). We know that velocity is the rate of change of position with respect to time. In this case, we want to find the velocity at x (subscript max), so we need to calculate the derivative of position (x) with respect to time (t):
v = dx / dt
To find the derivative of x with respect to t, we can differentiate the position function with respect to x and multiply it by the derivative of x with respect to t:
v = (dx / dt) = (dx / dt) / (dx / dx)
Now, let's find the derivative of x with respect to t. We know that x is a function of time, so we can write it as x(t). Taking the derivative, we have:
dx / dt = d(x(t)) / dt
To find x(t), we need to integrate the velocity function. Since we don't know the exact form of the force function F subscript x, we can't directly integrate it to obtain x(t). However, we can still calculate the velocity at x (subscript max) using our previous results.
At x (subscript max), we know that x = pi/2c, so we can substitute this value into our expression for velocity:
v = (dx / dt) / (dx / dx) = (dx / dt) / 1
Now, we just need to find dx / dt at x = pi/2c. To do this, we can differentiate the position function x with respect to time using the chain rule:
dx / dt = (dx / du) * (du / dt)
Here, we introduce a new variable u = cx, so we have:
dx / dt = (dx / du) * (du / dt) = (dx / du) * c
Now, let's differentiate x with respect to u:
dx / du = d(F subscript o sin (cx)) / du
Using the chain rule again, we find:
dx / du = F subscript o cos(cx) * c
Plugging this back into our expression for dx / dt:
dx / dt = (dx / du) * c = F subscript o cos(cx) * c
Now, we can evaluate this expression at x = pi/2c:
dx / dt = F subscript o cos((pi/2)c) * c
Since cos((pi/2)) = 0, we have:
dx / dt = 0
Therefore, the particle's velocity as it reaches x (subscript max) is zero.