A 0.25 kg particle moves along an x axis according to x(t) = - 10 + 3 t + 2 t2 - 5 t3, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.6 s ? Give an expression for the (a) x, (b) y and (c) z components.
force=mass*acceleration
= mass*d"(x)/dt"
= mass*d/dt (3+4t-15t^2)
= mass* (4-30t)
so put in t=3.6, and solve for force. Of course, it is in the x direction. the y component and z component is zero.
To find the net force acting on the particle at t = 3.6 s, we need to calculate the derivative of the position function x(t) twice to obtain the acceleration function, and then multiply the mass of the particle with the acceleration to get the net force.
First, let's find the expression for the x-coordinate component of the position x(t). Given:
x(t) = -10 + 3t + 2t^2 - 5t^3
To find the x-component, we can ignore the other variables and consider only the x-coordinate:
(a) x-component: x(t) = -10 + 3t + 2t^2 - 5t^3
Now, let's find the derivative of x(t) twice to get the acceleration function:
v(t) = dx(t)/dt = d/dt (-10 + 3t + 2t^2 - 5t^3)
= 3 + 4t - 15t^2
a(t) = dv(t)/dt = d/dt (3 + 4t - 15t^2)
= 4 - 30t
The expression for the x-component of the net force is obtained by multiplying the mass of the particle with the acceleration:
F_x = m * a(t)
= 0.25 kg * (4 - 30t)
Now let's move on to the y and z components.
Given that the motion is along the x-axis only, the particle's y and z-components remain at zero. Therefore, the expressions for the y and z components of the net force are both zero:
(b) y-component: F_y = 0
(c) z-component: F_z = 0
To summarize:
(a) x-component: F_x = 0.25 kg * (4 - 30t)
(b) y-component: F_y = 0
(c) z-component: F_z = 0
To find the net force acting on the particle, we need to calculate the second derivative of the position function with respect to time, which gives us the acceleration. Then, we can use Newton's second law, F = ma, to find the net force.
Let's start by finding the position function x(t) at t = 3.6 s.
x(t) = -10 + 3t + 2t^2 - 5t^3
Substituting t = 3.6 into x(t), we get:
x(3.6) = -10 + 3(3.6) + 2(3.6^2) - 5(3.6^3)
x(3.6) = -10 + 10.8 + 23.04 - 186.624
x(3.6) = -162.784
So, the x-coordinate of the particle at t = 3.6 s is -162.784 meters.
Next, let's find the velocity function v(t) by taking the first derivative of the position function with respect to time.
v(t) = d/dt(x(t))
v(t) = d/dt(-10 + 3t + 2t^2 - 5t^3)
v(t) = 3 + 4t - 15t^2
Now, let's find the velocity v(3.6) at t = 3.6 s.
v(3.6) = 3 + 4(3.6) - 15(3.6^2)
v(3.6) = 3 + 14.4 - 194.4
v(3.6) = -177
The velocity of the particle at t = 3.6 s is -177 m/s.
Finally, let's find the acceleration function a(t) by taking the second derivative of the position function with respect to time.
a(t) = d^2/dt^2(x(t))
a(t) = d^2/dt^2(-10 + 3t + 2t^2 - 5t^3)
a(t) = 4 - 30t
Now, let's find the acceleration a(3.6) at t = 3.6 s.
a(3.6) = 4 - 30(3.6)
a(3.6) = 4 - 108
a(3.6) = -104
The acceleration of the particle at t = 3.6 s is -104 m/s^2.
Since F = ma, the net force acting on the particle at t = 3.6 s is given as:
F = m * a
F = (0.25 kg) * (-104 m/s^2)
F = -26 N (in the negative x-direction)
Therefore, the x-component of the net force acting on the particle at t = 3.6 s is -26 N.
Since the problem statement does not provide any information about the y and z components of the force, we cannot determine their values and thus cannot express them in an equation.