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elena-14-01-66 [18.8K]
3 years ago
13

In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig

htless, what is the minimum input force needed to lift the weight? Rope Pile L= 16.74m
A. 61 N
B. 121 N
C. 181 N
D. 361 N
Physics
2 answers:
STALIN [3.7K]3 years ago
8 0
Your answer is D. 361 N
Sveta_85 [38]3 years ago
7 0

Answer:

A. 61 N

Explanation:

In this system, there are three moveable pulleys. In this case, the rope will have to move six times as far as the weight will rise. In other words, the input distance is 6 times the output distance  ( d in =6⋅ d out ) . Because the efficiency is 100%, the work input is equal to the work output, so we have:

work input  = work output

f( in)  ⋅  d( in) =   f( out)  ⋅  d( out)

 f( in)  ⋅ (6 ⋅  d( out))  =  f( out)  ⋅  d( out)

 f( in)  ⋅ 6 =  f( out)

  f( in)  ⋅ 6 = 360 N

  f( in)  = 60 N  

Another way to think of this is that since the rope has to move 6 times as far as the height of the weight, the mechanical advantage will be 6. Therefore, the input force needed is  360 ÷ 6 = 60 N . Of course, 60 N will balance the object, so a force slightly greater than 60 N (like 61 N) will be required to lift it.

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Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

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Where:

x = Final distance

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x = 1.66m      

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t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

x = x_{0} + V_{0}t + \frac{1}{2}at^2

1.66 = 0 + 0t + \frac{1}{2}9.8t^2

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\sqrt{0.339} = t\\ t = 0.582s

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

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