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N76 [4]
3 years ago
11

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

imb 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 θ = 60.6° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Physics
1 answer:
Anestetic [448]3 years ago
7 0

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

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Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

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Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

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On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

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We already know the value of  d and calculated a, we have to find t:

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t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

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If V_{o}=0

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V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

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V_{f}=2.96m/s    

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