The position of a 2.75×105n training helicopter under test is given by r⃗ =(0.020m/s3)t3i^+(2.2m/s)tj^−(0.060m/s2)t2k^. part a f

Question

Murljashka [212]

2 years ago

3

The position of a 2.75×105n training helicopter under test is given by r⃗ =(0.020m/s3)t3i^+(2.2m/s)tj^−(0.060m/s2)t2k^. part a f

ind the net force on the helicopter at t=5.0s.

Physics

2 answers:

Sophie [7] · Answer 1 · 2023-10-01T19:33:23+03:00

The net force on the helicopter at time $t = 5.0\,{\text{s}}$ is $\fbox{\begin\\\left( {1.68 \times {{10}^4}\,\hat i - 3.36 \times {{10}^3}\,\hat j} \right)\,{\text{N}}\end{minispace}}$ .

Further Explanation:

The helicopter moves under the action of the force which changes the position of the helicopter with respect to time according to the following equation.

$\vec{r}=(0.020\text{ m}/\text{s}^3)\text{ t}^3\hat{i}+(2.2\text{ m}/\text{s})\text{ t}^3\hat{j}-(0.060\text{ m}/\text{s}^2)\text{ t}^2\hat{k}$

The force acting on the helicopter at a particular time is given by:

$\fbox{\begin\\F= m\cdot{a}\end{minispace}}$

Here, $m$ is the mass of the helicopter and $a$ is the acceleration of helicopter.

The helicopter has a weight of $2.75 \times {10^5}\,{\text{N}}$ . This weight of the helicopter can be converted into the mass of the helicopter.

The weight of the helicopter can be represented as.

$\fbox{\begin\\W=m \cdot g\end{minispace}}$

Rearrange and simplify the expression for the mass of helicopter.

$\begin{aligned}m&=\frac{W}{g}\\&=\frac{{2.75\times{{10}^5}}}{{9.8}}\,{\text{kg}}\\&=2.{\text{80}}\times{\text{1}}{{\text{0}}^4}\,{\text{kg}}\\\end{aligned}$

The velocity of the helicopter can be determined by the differentiation of the position vector of the helicopter with respect to time.

$\begin{aligned}\vec{v}&=\dfrac{{d\vec{r}}}{{dt}}\\&=\dfrac{d}{{dt}}\ [(0.020\text{ m}/\text{s}^3)\text{ t}^3\hat{i}+(2.2\text{ m}/\text{s})\text{ t}^3\hat{j}-(0.060\text{ m}/\text{s}^2)\text{ t}^2\hat{k}]\\&=0.060{t^2}\hat{i}+2.2\hat{j}-0.12t\hat{k}\end{aligned}$

The acceleration of the helicopter is the derivative of the velocity with respect to the time.

$\begin{aligned}\vec a&=\frac{{d\vec v}}{{dt}}\\&=\frac{d}{{dt}}\left({0.060{t^2}\hat i+2.2\hat j-0.12t\hat k}\right)\\&=0.12t\hat i-0.12\hat k\\\end{aligned}$

Substitute the value of $t = 5.0\,{\text{s}}$ in equation of the acceleration.

$\begin{aligned}a&=0.12\left({5.0}\right)\hat i-0.12\hat k\\&=0.6\hat i-0.12\hat k\\\end{aligned}$

Substitute the values of the acceleration and the mass of the helicopter in the equation of force.

$\begin{aligned}F&=\left( {2.80\times{{10}^4}\,{\text{kg}}}\right)\left({0.6\hat i-0.12\hat k}\right) \\&=\left({1.68\times{{10}^4}\,\hat i-3.36\times{{10}^3}\,\hat j}\right)\,{\text{N}}\\\end{aligned}$

Thus, the net force acting on the helicopter at time $t = 5\,{\text{s}}$ is $\fbox{\begin\\\left( {1.68 \times {{10}^4}\,\hat i - 3.36 \times {{10}^3}\,\hat j} \right)\,{\text{N}}\end{minispace}}$ .

Learn More:

1. The motion of a body under friction brainly.com/question/7031524

2. Conservation of energy brainly.com/question/3943029

3. Motion in two dimensions brainly.com/question/11023695

Answer Details:

Grade: College

Subject: Physics

Chapter: Force and acceleration

Keywords:

Helicopter, force, time, 5s, position, velocity, derivative, acceleration, weight, net force, t=5s, 0.020t3 m/s3, 2.75x10^5 N, helicopter under test, gravity, mass of helicopter.

Sergio039 [100] · Answer 2 · 2023-10-01T17:35:12+03:00

<em>The net force on the helicopter at t=5.0s = F = 1.65.10⁴i - 3.3.10³k </em>

<h3><em>Further explanation </em></h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

$\large{\boxed{\bold{a~=~\frac{F}{m} }}}$