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2 years ago

A 75.0 kg sailor climbs a 28.3 m rope ladder at and angle of 45.0degrees with the mast. how much work did he do?

2 answers:

8 0

W = m · g · h
h = 28.3 m · sin 45° = 28.3 m · 0.707 = 20 m
g = 9.8 m/s²
W = 75 kg · 9.8 m/s² · 20 m
Answer:
W = 14,700 J = 14.7 kJ

crimeas [40]2 years ago

7 0

Explanation :

It is given that,

mass of sailor, m = 75 Kg

length of rope, h = 28.3 m

work done, $w=Fdcos\theta$

$w=mghcos\theta$

$w=75\ Kg\times 9.8\ m/s^2\times 28.3\ m\ cos(45)$

$w=10,816\ J$

$w=10.816\ kJ$

Hence, it is the required solution.

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