Question

A 0.340 kg particle moves in an xy plane according to x(t) = −15.00 + 2.00t − 4.00t3 and y(t) = 25.00 + 7.00t − 9.00t2, with x and y in meters and t in seconds. At t = 0.700 s, what are (a) the magnitude and (b) the angle (relative to the positive direction of the x axis) of the net force on the particle?

Answer 1

Answer:F=8.37 N

Explanation:

Given

mass of Particle $m=0.34 kg$

Position of Particel $x(t)=-15 +2t-4t^3$

$y(t)=25+7t-9t^2$

velocity is given by $\frac{\mathrm{d} x}{\mathrm{d} t}=v_x$

$v_x=\frac{\mathrm{d} x}{\mathrm{d} t}$

and acceleration is given by

$a_x=\frac{\mathrm{d^2} x}{\mathrm{d} t^2}$

$a_x=-4\times 3\times 2t$

at $t=0.7 s$

$a_x=-24 \cdot 0.7=16.8 m/s^2$

Similarly acceleration in y direction is given by

$a_y=-9\times 2=-18 m/s^2$

net acceleration is given by

$a_{net}=\sqrt{a_x^2+a_y^2}$

$a_{net}=\sqrt{16.8^2+18^2}$

$a_{net}=24.62 m/s^2$

Therefore net Force $F=m\cdot a_{net}$

$F=0.34\cdot 24.62=8.37 N$

angle which Force makes with horizontal is given by

$\tan \theta =\frac{a_y}{a_x}=\frac{-18}{-24t}$

$\tan \theta =\frac{3}{4t}$

at $t=0.7 s$

$\tan \theta =\frac{3}{2.8}$

$\theta =46.97^{\circ}$

Answer 2

(a) The magnitude of the net force exerted by the given particle is 8.37 N.

(b) The angle made by the net force with the x-axis is obtained as $46.97^\circ$.

Newton's Second Law of Motion

(a) It is given that the particle has a mass of $m = 0.340\,kg$.

Also, the x-component of the position is given by,

$x(t)= -15+2t-4t^3$

So, the x-component of acceleration is given  by,

$a_x = \frac{dv_x}{dt} =\frac{d^2 x}{dt^2}$

$\implies a_x = \frac{d^2}{dt^2} (-15+2t-4t^3)=\frac{d}{dt} (2-12t^2)= -24t$

But given that $t = 0.7\,s$. Therefore, we get;

$a_x = -24\times 0.7=-16.8\,m/s^2$

The y-component of the position is given by,

$y(t)= 25+7t-9t^2$

So, the y-component of acceleration is given  by,

$a_y = \frac{dv_y}{dt} = \frac{d^2 x_y}{dt^2}$

$\implies a_y = \frac{d^2}{dt^2} (25+7t-9t^2)=\frac{d}{dt} (7-18t)= -18\,m/s^2$

Therefore, the magnitude of the net acceleration is given by;

$a_{net}=\sqrt{(a_x)^2 + (a_y)^2\,} =\sqrt{(-16.8\,m/s^2)^2 + (-18\,m/s^2)^2} =24.62\,m/s^2$

So, the magnitude of the net force can be calculated using Newton's second law of motion.

$F_{net}=ma_{net}$

Substituting the known values, we get;

$F_{net}=(0.340\,kg)\times (24.62\,m/s^2)=8.37\,N$

(b) The angle made by the net force vector with the horizontal can be found out by calculating the angle formed by the acceleration vector.

$tan\, \theta=\frac{a_y}{a_x} =\frac{-18\,m/s^2}{-16.8\,m/s^2} =1.0714$

$\implies \theta = tan^{-1}(1.0714)=46.97^\circ$

Learn more about Newton's second law of motion here:

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