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3 years ago

the net force on the propeller of a 3.2 kg model is 7.0 N forward. what is the acceleration of the plane

Physics

2 answers:

erica [24]3 years ago

7 0

Answer:

Acceleration of the plane, $a=2.18\ m/s^2$

Explanation:

It is given that,

Mass of the model, m = 3.2 kg

The net force acting on the propeller, F = 7 N

We need to find the acceleration of the plane. Let it is given by a. Using second law of motion, we can find the value of a. The mathematical relationship between the mass, force and the acceleration is given by :

$F=m\times a$

$a=\dfrac{F}{m}$

$a=\dfrac{7\ N}{3.2\ kg}$

$a=2.18\ m/s^2$

So, the acceleration of the plane is $2.18\ m/s^2$ . Hence, this is the required solution.

Masja [62]3 years ago

3 0

The acceleration is 2.19m/s^2

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3 years ago

Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at <em>t</em> = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago