A car goes North a distance of 120 kilometers during a time period of 3.0 hours. What is the average
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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 = 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