The position of a 2.75 *10^5 N training helicopter under test is given by
r = (0.020 m/s^3)(t^3)i + (2.2 m/s)(t)j - (0.060 m/s^2)(t^2) k
Find the net force on the helicopter at t =5.0s.
Express your answer in terms of i,j,k. Express your coefficient using two significant figures.
just plug in 5 for t:
r = (0.020 m/s^3)(5^3)i + (2.2 m/s)(5)j - (0.060 m/s^2)(5^2) k
1. take the second derivative of r to find acceleration.
2. then divide 2.75 *10^5 N by 9.8m/s^2 to find how much the helicopter weighs in kg
3. multiply your result from 1 with your result from 2
To find the net force on the helicopter, we need to calculate its acceleration at t = 5.0s and then multiply it by its mass.
Given position function:
r = (0.020 m/s^3)(t^3)i + (2.2 m/s)(t)j - (0.060 m/s^2)(t^2)k
To find acceleration, we need to differentiate the position function twice with respect to time (t).
First derivative of r with respect to t (dr/dt):
dr/dt = (d/dt)[(0.020 m/s^3)(t^3)i + (2.2 m/s)(t)j - (0.060 m/s^2)(t^2)k]
= (0.060 m/s^2)(t^2)i + (2.2 m/s)j - (0.120 m/s^2)(t)k
Second derivative of r with respect to t (d^2r/dt^2):
(d^2r/dt^2) = (d/dt)[(0.060 m/s^2)(t^2)i + (2.2 m/s)j - (0.120 m/s^2)(t)k]
= (0.120 m/s^2)(t)i - (0.120 m/s^2)k
Now we can calculate the net force F using Newton's second law, F = m * a.
Given the force F = 2.75 * 10^5 N,
F = m * (d^2r/dt^2)
Plugging in the given force and expressing it in terms of i, j, and k:
2.75 * 10^5 N = m * ((0.120 m/s^2)(t)i - (0.120 m/s^2)k)
At t = 5.0s, the force F can be written as:
2.75 * 10^5 N = m * ((0.120 m/s^2)(5.0)i - (0.120 m/s^2)k)
To express the coefficient using two significant figures, we can round it to 0.12:
m * (0.12 m/s^2)(5.0)i - (0.12 m/s^2)k
Therefore, the net force on the helicopter at t = 5.0s is approximately:
(0.12 m/s^2)(5.0)i - (0.12 m/s^2)k, where the coefficient is rounded to two significant figures.
To find the net force on the helicopter at t = 5.0s, we need to calculate the derivative of the position vector with respect to time and then substitute the value of t into the resulting expression.
Given that the position of the helicopter is given by the vector:
r = (0.020 m/s³)(t³)i + (2.2 m/s)(t)j - (0.060 m/s²)(t²)k
Let's find the derivative of the position vector with respect to time:
dr/dt = (0.020 m/s³)(3t²)i + (2.2 m/s)j - (0.060 m/s²)(2t)k
Now, substitute t = 5.0s into the expression:
dr/dt = (0.020 m/s³)(3(5.0s)²)i + (2.2 m/s)j - (0.060 m/s²)(2(5.0s))k
dr/dt = (0.020 m/s³)(3(25.0s²)i + (2.2 m/s)j - (0.060 m/s²)(10.0s)k
dr/dt = (0.020 m/s³)(75.0s²)i + (2.2 m/s)j - (0.060 m/s²)(10.0s)k
dr/dt = (0.020 m/s³ * 75.0 * 10⁻³ * s²)i + (2.2 m/s)j - (0.060 m/s² * 10.0 * s)k
dr/dt = 0.00150i + 2.2j - 0.60k
The derivative of the position vector with respect to time gives us the velocity vector (v) at any given time.
Now, to find the net force on the helicopter, we multiply the mass of the helicopter (m) by the acceleration vector (a).
Considering that net force (F_net) is equal to mass (m) multiplied by acceleration (a), we can write:
F_net = m * a
Since the mass of the helicopter is not given, we cannot calculate the net force. We would need to know the mass in order to calculate it.
However, if we were given a mass value, we could multiply it by the acceleration vector obtained from the derivative to find the net force on the helicopter.