Spiderman, whose mass is 75.0 kg, is dangling on the free end of a L = 11.5-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to sswing enough that he can reach a ledge when the rope makes a 59.3° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver? Part 1 of 4 - Conceptualize The force of the Earth on the man is fairly large and tends to slow his upward motion. Considering the mass involved and the value of g, we estimate that gravity should do hundreds, possibly thousands of joules of negative work on him. A rope of length L makes an angle of 6 with the vertical, and a figure of a man clings to the end of the rope. A vertical line labeled L shows how the rope would look if it were oriented vertically. The vertical distance between the bottom of this vertical line and the bottom of the angled rope is labeled h.

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Spiderman, whose mass is 75.0 kg, is dangling on the free end of a L = 11.5-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist,
he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a 59.3° angle with the vertical. How much work was
done by the gravitational force on Spiderman in this maneuver?
Part 1 of 4 - Conceptualize
The force of the Earth on the man is fairly large and tends to slow his upward motion. Considering the mass involved and the value of q, we estimate that gravity should do
hundreds, possibly thousands of joules of negative work on him.
A rope of length L makes an angle of 0 with the vertical, and a figure of a man clings to the end of the rope. A vertical line labeled L shows how the rope would look if
it were oriented vertically. The vertical distance between the bottom of this vertical line and the bottom of the angled rope is labeled h.
Part 2 of 4 - Categorize
He swings through several or many cycles, each going farther than the last. Gravity does positive work on him in each downswing and negative work in each upswing. we can use
the definition of work by a constant force between the original starting point and the final endpoint.
Part 3 of 4 - Analyze
The work done is W = F· Ar, where the gravitational force on Spiderman is
F, = -mg)
| 75 kg)
-9.80 m/s2 )j =
= -735
-735 jN.
75
He travels through net displacement given by the following.
Ar = L sin 59.3° î +
1- cos
59.3 •
11.5 m)sin
59.3 oî +
11.5 m)(1
59.3 •i
- cos
Transcribed Image Text:Spiderman, whose mass is 75.0 kg, is dangling on the free end of a L = 11.5-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a 59.3° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver? Part 1 of 4 - Conceptualize The force of the Earth on the man is fairly large and tends to slow his upward motion. Considering the mass involved and the value of q, we estimate that gravity should do hundreds, possibly thousands of joules of negative work on him. A rope of length L makes an angle of 0 with the vertical, and a figure of a man clings to the end of the rope. A vertical line labeled L shows how the rope would look if it were oriented vertically. The vertical distance between the bottom of this vertical line and the bottom of the angled rope is labeled h. Part 2 of 4 - Categorize He swings through several or many cycles, each going farther than the last. Gravity does positive work on him in each downswing and negative work in each upswing. we can use the definition of work by a constant force between the original starting point and the final endpoint. Part 3 of 4 - Analyze The work done is W = F· Ar, where the gravitational force on Spiderman is F, = -mg) | 75 kg) -9.80 m/s2 )j = = -735 -735 jN. 75 He travels through net displacement given by the following. Ar = L sin 59.3° î + 1- cos 59.3 • 11.5 m)sin 59.3 oî + 11.5 m)(1 59.3 •i - cos
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