a) [tex]F=(3675i-4543k)N[/tex]
b) 5843 N
Explanation:
a)
The position of the UFO at time t is given by the vector:
[tex]r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k[/tex]
Therefore it has 3 components:
[tex]r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2[/tex]
We start by finding the velocity of the UFO, which is given by the derivative of the position:
[tex]v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t[/tex]
And then, by differentiating again, we find the acceleration:
[tex]a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6[/tex]
The weight of the UFO is W = 12,500 N, so its mass is:
[tex]m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg[/tex]
Therefore, the components of the force on the UFO are given by Newton's second law:
[tex]F=ma[/tex]
So, Substituting t = 2 s, we find:
[tex]F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N[/tex]
So the net force on the UFO at t = 2 s is
[tex]F=(3675i-4543k)N[/tex]
b)
The magnitude of a 3-dimensional vector is given by
[tex]|v|=\sqrt{v_x^2+v_y^2+v_z^2}[/tex]
where
[tex]v_x,v_y,v_z[/tex] are the three components of the vector
In this problem, the three components of the net force are:
[tex]F_x=3675 N\\F_y=0\\F_z=-4543 N[/tex]
Therefore, substituting into the equation, we find the magnitude of the net force:
[tex]|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N[/tex]
