physics and mathematics

Question

A force applied to a small ball with the mass m = 1

(kg) is: F= y.e_x-x.e_y+z.e_z

The ball follows the trajectory: r(t)= t.e_x+e_y-t.e_z

Find the work W produced by the force F along the given trajectory for the first 10 seconds.

Check if the force F is potential.

Is the given force F a net force? Give your arguments.

Show F and r(t) in matrix form independently.

Answer 1

Let me rewrite the given information using, what I speculate, is meant by "matrix form".

F(x,y,z) = <y, -x, z>,

r(t) = <t, 1, -t>

FIND THE WORK:

By definition of work, W is the definite integral over the distance travelled by the ball

in the first 10 seconds

W = ∫F.dr (1)

We will convert that to an integral over time.

The derivative

dr/dt = <1, 0, -1>

So

dr = <1, 0, -1> dt

At a time instant t the force acting on the ball is obtained by substituting the

position of the ball at the time t into the coordinates in the definition of F.

That is, substitute

x(t) = t,

y(t) = 1,

z(t) = -t

into F(x, y, z) resulting in

F(t) = <1, -t, -t>

Plugging both F(t) and dr into (1) results in the definite integral over the time period 0 to 10 seconds:

W = ∫<1, -t, -t>.<1, 0, -1>dt

= ∫(1 + t)dt

= (t + t²/2) between 0 and 10

= 10 + 10²/2 = 60

IS THE FORCE POTENTIAL?

The force is potential iff its curl is 0.

curl(F(x,y,z)) = <∂Fz/∂y - ∂Fy/∂z, ∂Fx/∂z - ∂Fz/∂x, ∂Fy/∂x - ∂Fx/∂y>

= <0, 0, -2>

Therefore the force is not potential.

IS F A NET FORCE?

By Newton's Second Law the net force is

m d²r/dt² = 1.d²/dt²<t, 1, -t> = d/dt<1, 0, -1> = <0, 0, 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,

therefore subject to no net force.