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Calculus/ physics question HELP!!! thanks?
Suppose that 6 J of work is needed to stretch a spring from its natural length of 24 cm to a length of 36 cm.
(a) How much work is needed to stretch the spring from 29 cm to 34 cm? (Round your answer to two decimal places.)
my answer is 6.25 (25/4) and it says its incorrect... please help thank you
Incorrect: Your answer is incorrect.
4 Answers
- PinkgreenLv 74 years ago
F=kx=>
x..........x
SFdx=Skxdx=>
0.........0
w=k(x^2)/2
where k=constant;
x=the net increment in length;
F=the force applied
w=the work done needed to
stretch the length by x.
When W=6 J,
x=36-24=12 cm=0.12 m
=>k=2500/3
x=29-24=5cm=0.05m
=>
w(0.05)=(2500/3)(0.05^2)/2
=>
w(0.05)=1.0417 J
x=34-24=10 cm=0.1 m
=>
w(0.1)=(2500/3)(0.1^2)/2
=>
w(0.1)=4.1666 J
Ans. it needs w(o.1)-w(0.05)=
3.12 J approximately to stretch
the spring from 29 cm to 34 cm.
- RealProLv 74 years ago
How do you expect it to be more than 6 J when it starts further than 24 and stretches to less than 36? Just sayin.
- AshLv 74 years ago
W = ½kx²
6 = ½k(36-24)²
k= 12/12²
k = 1/12
W(29 to 34) = W(34) - W(29)
W(29 to 34) = ½(1/12)(34-24)² - ½(1/12)(29-24)²
W(29 to 34) = (1/24)[(10)²-(5)²]
W(29 to 34) = 3.125 J