A 0.41 kg particle moves in an xy plane according to x(t) = - 16 + 1 t - 3 t3 and y(t) = 22 + 3 t - 10 t2, with x and y in meters and t in seconds. At t = 1.8 s,

Question

what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?

Answer 1

The position of the particle is given by the equations:

x(t) = -16 + t - 3t^3 y(t) = 22 + 3t - 10t^2

Taking the derivatives with respect to time, we get:

vx(t) = 1 - 9t^2 vy(t) = 3 - 20t

At t = 1.8 s, we have:

vx(1.8) = 1 - 9(1.8)^2 = -29.16 m/s vy(1.8) = 3 - 20(1.8) = -33 m/s

The acceleration of the particle is given by the second derivative of the position with respect to time:

ax(t) = -18t ay(t) = -20

At t = 1.8 s, we have:

ax(1.8) = -32.4 m/s^2 ay(1.8) = -20 m/s^2

The net force on the particle is given by:

Fnet = ma

where m is the mass of the particle.

Substituting the given values, we get:

Fnet = 0.41 kg × √((-32.4)^2 + (-20)^2) = 20.11 N

The magnitude of the net force on the particle is approximately 20.11 N.

The angle of the net force relative to the positive direction of the x-axis is given by:

θ = tan^-1(Fy/Fx)

where Fx and Fy are the x and y components of the net force.

Substituting the given values, we get:

θ = tan^-1(-33/(-29.16)) = -48.9°

Therefore, the angle of the net force on the particle is approximately -48.9°.

The angle of the particle's direction of travel at t = 1.8 s is given by:

φ = tan^-1(vy/vx)

Substituting the given values, we get:

φ = tan^-1(-33/(-29.16)) = 47.6°

Therefore, the angle of the particle's direction of travel at t = 1.8 s is approximately 47.6°.