Work and spring constant questions?

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To determine the depth of a well from which an 11.0 kg bucket is lifted with 7.00 kJ of work, the force of gravity (9.81 m/s²) must be used to calculate the required force, leading to a depth of approximately 64.87 meters. For the spring problem, Hooke's law is applied to find the spring constant when a 7.60 kg object stretches the spring from 36.0 cm to 45.50 cm. The spring's length when pulled by two people with 170 N each can also be calculated using the appropriate equations for force and displacement. The discussions highlight the importance of correctly applying physics formulas to solve work and spring constant problems. Understanding these concepts is crucial for success in physics assessments.
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1) If a man lifts a 11.0 kg bucket from a well and does 7.00 kJ of work, how deep is the well? Assume that the speed of the bucket remains constant as it is lifted.

2) Hooke's law describes a certain light spring of unstretched length 36.0 cm. When one end is attached to the top of a door frame, and a 7.60-kg object is hung from the other end, the length of the spring is 45.50 cm.

(a) Find its spring constant.
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 170 N. Find the length of the spring in this situation.

work=fd
PE=(1/2)k(x^2)

ok for the first one i did 7000J/11= 636.36 and it says that's wrong
and for the second one I am not even sure where to start
 
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bbanas0695 said:
1) If a man lifts a 11.0 kg bucket from a well and does 7.00 kJ of work, how deep is the well? Assume that the speed of the bucket remains constant as it is lifted.

work=fd


ok for the first one i did 7000J/11= 636.36 and it says that's wrong

You are using the correct equation, work = fd. So, what is f, the force required to lift a mass of 11 kg? Hint: it is the same as the force of gravity exerted on the mass.

2) Hooke's law describes a certain light spring of unstretched length 36.0 cm. When one end is attached to the top of a door frame, and a 7.60-kg object is hung from the other end, the length of the spring is 45.50 cm.

(a) Find its spring constant.
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 170 N. Find the length of the spring in this situation.


PE=(1/2)k(x^2)
There is another useful equation for springs, relating force and displacement (amount of stretch) of the end of the spring. Find that equation and try to apply it here.
 
You are using the correct equation, work = fd. So, what is f, the force required to lift a mass of 11 kg? Hint: it is the same as the force of gravity exerted on the mass.

the force of gravity is 9.81 so that would be the force. so 7000/9.81= 713.56 m ?

and for the second one, i guess we don't have to do it. the answer in the website is wrong or something.
 
oh wait. you need to find force using f=ma so 11*9.81=107.91
and then plug that in which would be 7000/107.91= 64.87m
i already got it wrong on the homework but thanks. At least I know how to do it now :)
 
bbanas0695 said:
oh wait. you need to find force using f=ma so 11*9.81=107.91
and then plug that in which would be 7000/107.91= 64.87m
i already got it wrong on the homework but thanks. At least I know how to do it now :)
Looks good. Too late for the homework, but remember it for the test :smile:
 
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