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 $t=0.7 s$

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

$\theta =46.97^{\circ}$