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A 750.0-kg boulder is raised from a quarry 125 m deep by a long uniform chain having a mass of 575 kg. this chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) what is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it tak

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meerkat18
For this problem, we apply Newton's Second Law of Motion. This law states that the net force of a system is equal to the product of the mass of the total system and acceleration. In equation, it is written as

F = ma

Part A. The net force in this motion in a vertical axis is the difference between the tension of the chain and the mass of the object. The tension is 2.5 times the weight of the chain: T = 2.5*(575 kg+750 kg) * 9.81 m/s² = 32,495.625 N. Applying the equation,

T - mg = ma
32,495.625 - (750)(9.81) = (750)(a)
a = 33.52 m/s²

For the part B, I can't answer the question because I would need other data like the height of the object it travelled.

Answer:

Part a)

[tex]a = 0.82 m/s^2[/tex]

Part b)

[tex]t = 17.44 s[/tex]

Explanation:

Part a)

Maximum tension that is obtained by the rope is 2.5 times the weight of the rope

So here we know that

mass of the boulder = 750 kg

mass of the chain = 575 kg

now total mass of the boulder and chain = 750 kg + 575 kg

[tex]M = 1325 kg[/tex]

Now we have maximum tension in the rope is given as

[tex]T = 2.5 \times 575 \times 9.8[/tex]

[tex]T = 14087.5 N[/tex]

now from Newton's 2nd law we have

[tex]T - Mg = Ma[/tex]

[tex]14087.5 - (1325\times 9.81) = 1325 a[/tex]

[tex]a = 0.82 m/s^2[/tex]

Part b)

Time taken by the boulder to raise by 125 m depth

we can use kinematics and we will assume here that it will start from rest and accelerate at above maximum value that we found in part a)

so by kinematics

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]125 = 0 + \frac{1}{2}(0.82)t^2[/tex]

[tex]t = 17.44 s[/tex]

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