Question

1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is <0, , 0>, where is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the direction towards where the ball lands, and the direction taken to be upwards.)

Answer 1

Answer:

d = <23, 33, 0> m ,    F_W = <0, -9.8, 0> ,   W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

         x = vox t

Y Axis  

         y = $v_{oy}$ t - ½ g t²

It's displacement is

      d = x i ^ + y j ^ + z k ^

Substituting

      d = (23 i ^ + 33 j ^ + 0) m

Using your notation

   d = <23, 33, 0> m

The force of gravity is the weight of the body

         W = m g

        W = 1  9.8 = 9.8 N

In vector notation, in general the upward direction is positive

         W = (0 i ^  - 9.8 j ^ + 0K ^) N

         W = <0, -9.8, 0>

Work is defined

           W = F. dy

             W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

          Cos 180 = -1

           W = -F y

           W = - 9.8 (33-0)

           W = -323.4 J