Answer:
The woman's force was directed 59.22⁰ to the horizontal.
Explanation:
Given;
work done by the woman, W = 236 J
distance through the load was moved, d = 24.4 m
applied force, F = 18.9 N
inclination of the force, = θ
The work done by the woman is calculated as;
W = Fdcosθ
[tex]cos \ \theta = \frac{W}{Fd} \\\\cos \ \theta = \frac{236 }{18.9 \times 24.4} \\\\cos \ \theta = 0.5118\\\\\theta = cos^{-1} ( 0.5118)\\\\\theta =59.22^0[/tex]
Therefore, the woman's force was directed 59.22⁰ to the horizontal.
