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Finding the Work done on a Ride? 
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#1
Jul1210, 09:28 PM

P: 24

1. The problem statement, all variables and given/known data
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? 2. Relevant equations W = FsCos I don't know if this is the right formula to use! 3. 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? 


#2
Jul1210, 09:39 PM

P: 452

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.



#3
Jul1210, 09:56 PM

P: 24

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


#4
Jul1210, 10:00 PM

P: 452

Finding the Work done on a Ride?



#5
Jul1210, 10:02 PM

P: 24

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? 


#6
Jul1210, 10:18 PM

P: 452

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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