Finding the Work done on a Ride?

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Homework Help Overview

The problem involves calculating the work done on a child being lifted on a ride, specifically the Batwing ride at Six Flags America. The scenario includes a lift incline of 66.2 m at an angle of 32.0 degrees, with the child weighing 43 kg. The original poster expresses uncertainty about the appropriate formula for calculating work.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of the formula W = FsCos and question whether it is the correct approach. There is mention of using a free body diagram and considering energy methods. Some participants explore different trigonometric functions (sine vs cosine) in the context of calculating work.

Discussion Status

The discussion is ongoing, with participants providing guidance on the use of diagrams and the importance of understanding the direction of forces. There is an exploration of different interpretations regarding the angle used in calculations, but no consensus has been reached yet.

Contextual Notes

Participants note confusion regarding the application of trigonometric functions in the context of work and the need to clarify the definitions and directions involved in the problem.

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Finding the Work done on a Ride?

Homework Statement



The lift incline on the Batwing ride at Six Flags America, is 66.2 m long and has an incline of 32.0 degrees. How much work is done to take a 43 Kg child to the top of the hill?

Homework Equations


W = FsCos
I don't know if this is the right formula to use!

The Attempt at a Solution


If this is the right formula to use would the solution go like this:
F = 43 Kg x 9.81 m/s/s
F = 421.83N

W = FsCos
W = 421.83N x 66.2 m x (Cos 32)
W = 23681.8669
W = 24000 J

Is this right, or did I do this wrong? If I did do this wrong can someone help me?
 
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You should do a free body diagram. You have a problem in how You chose your angle or do your trig.You can also do the problem using energy methods.
 


wait wait what if I did this instead?
W = 421.83N x 66.2 m x (Sin 32)
 


ilovemynny said:
wait wait what if I did this instead?
W = 421.83N x 66.2 m x (Sin 32)

This is correct but do you understand why? Can you solve it using energy?
 


actually no >.> my previous questions used the sin when finding height
but when i looked up the formula for work with angles it gave me cos.
can you tell me why?
 


ilovemynny said:
actually no >.> my previous questions used the sin when finding height
but when i looked up the formula for work with angles it gave me cos.
can you tell me why?
Ok. W=F* cos(a)*s is the definition of work when we have an abject moving in a straight line. It means that work is equal to the component of the Force acting on the same direction as the objects movement.If you have force acting in the same direction as the object moves the angle is 0 and W=Fs if the force acts at a angle W=F* cos(a)*s.The thing is that you don't use this formula to plug random things in it.You must figure out what is the displacement "s", what is the direction of the objects movement and at what angle "a" does the force does with that direction. You do that by making a Free Body diagram.
 

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