F = y.ex – x.ey + z.ez is the force exerted to a tiny ball with mass m = 1 (kg). The trajectory of the ball is as follows: t.ex + ey – t.ez = r(t).

Question

F = y.ex – x.ey + z.ez is the force exerted to a tiny ball with mass m = 1 (kg). The trajectory of the ball is as follows: t.ex + ey – t.ez = r(t).

a) For the first 10 seconds, calculate the work W produced by the force F along the given trajectory.

b) Determine if the force F is potential.

c) Is the force F given a net force? Explain your arguments

Answer 1

I will speculate that your notation means

F(t) = y(t).e_x - x(t).e_y + z(t).e_z

where e_x is the unit vector in the x-direction,

e_y is the unit vector in the y-direction, and

e_z is the unit vector in the z-direction.

And (x, y, z) is the position vector of the ball, given by

(x(t), y(t), z(t)) = r(t) = t.e_x + e_y - t.e_z

Plugging into F(x,y,z):

F(t) = e_x - t.e_y - t.e_z

a)

By definition of work, W is the definite integral from 0 to 10

W = ∫F.dr (1)

The derivative

dr/dt = e_x - e_z

So

dr = (e_x - e_z)dt

Plugging both F and dr into (1)

W = ∫(e_x - t.e_y - t.e_z).(e_x - e_z)dt

= ∫(1 + t)dt

= (t + t²/2) between 0 and 10

= 10 + 10²/2 = 60 J

b)

The force is potential iff its curl is 0.

curl(F(x,y,z)) = (∂F_z/∂y - ∂F_y/∂z)e_x + (∂F_x/∂z - ∂F_z/∂x)e_y + (∂F_y/∂x - ∂F_x/∂y)e_z

= (0 - 0)e_x + (0 - 0)e_y + (-1 - 1)e_z = -2e_z

Therefore the force is not potential.

c)

By Newton's Second Law the net force is

md²r/dt² = 1.d²/dt²(t.e_x + e_y - t.e_z) = d/dt(e_x - e_z) = 0

The net force is 0, which is different from F.

Actually that was quite clear at the outset, because r(t) is a straight line with constant velocity.