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

madreJ

2 months 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:

Sav [3.1K] · Answer 1 · 2026-02-16T17:34:41+03:00

At the moment $t = 5.0\,{\text{s}}$ , the helicopter experiences a net force of $\fbox{\begin\\\left( {1.68 \times {{10}^4}\,\hat i - 3.36 \times {{10}^3}\,\hat j} \right)\,{\text{N}}\end{minispace}}$ .

Detailed Explanation:

The helicopter's motion results from a force that alters its position over time, governed by the equation below.

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

The force applied to the helicopter at any given time is expressed as:

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

Where $m$ denotes the helicopter's mass and $a$ indicates its acceleration.

The helicopter's weight is $2.75 \times {10^5}\,{\text{N}}$ , which can be used to calculate its mass.

The weight equation for the helicopter is:

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

Rearranging the formula allows simplification to obtain the helicopter's mass:

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

The helicopter's velocity is found by differentiating its position vector 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 corresponds to the time derivative of the velocity.

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

By substituting $t = 5.0\,{\text{s}}$ into the acceleration formula, we get:

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

Replacing both the acceleration and mass values into the force equation yields:

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

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

serg [3.5K] · Answer 2 · 2026-02-16T17:35:01+03:00

The helicopter's net force at t = 5.0 seconds is F = 1.65 × 10⁴ i - 3.3 × 10³ k

Additional explanation

Newton's second law states that the acceleration caused by the net force acting on an object is directly proportional to that force and inversely proportional to the object's mass.

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