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?
Please help!!!
I don't think I can help you with "format errors" on the particular program you must be using.
To find the particle's velocity as it reaches x (subscript max), we can use the work-energy principle. This principle states that the work done by all forces on a particle is equal to the change in its kinetic energy.
In this case, the only force acting on the particle is the variable force F subscript x = F subscript o sin (cx). So, the work done by this force can be calculated by integrating the force with respect to displacement from x = 0 to x (subscript max).
To integrate the force, recall that the integral of sin(cx) with respect to x is - (1/c) cos(cx).
So, the work done by the force to move the particle from x = 0 to x (subscript max) is given by:
W = ∫ [F subscript o sin(cx)] dx
= - (F subscript o / c) ∫ cos(cx) dx
Integrating the right side, we get:
W = - (F subscript o / c) [sin(cx) / c]
The change in kinetic energy of the particle is equal to the work done by the force:
ΔKE = W
Since the particle starts from rest (V subscript o = 0 m/s), the initial kinetic energy is zero. Therefore, ΔKE is equal to the final kinetic energy:
ΔKE = (1/2) m v^2
Setting the work done equal to the change in kinetic energy:
- (F subscript o / c) [sin(cx) / c] = (1/2) m v^2
Simplifying the equation, we get:
v^2 = -2(F subscript o / c^2) [sin(cx) / c]
Taking the square root of both sides:
v = ± (√[-2(F subscript o / c^2) [sin(cx) / c]])
Since the particle is moving to the right along the x-axis, we can disregard the negative sign. Therefore, the particle's velocity as it reaches x (subscript max) is given by:
v = √[-2(F subscript o / c^2) [sin(cx) / c]]