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Question Number 125048 by bemath last updated on 08/Dec/20

It takes a force of 19,000 lb to  compress a spring from its free  height of 15 in to its fully   compressed height of 10 in. How  much  work does it take to   compress the spring the first in?  (a) 1900 in.−lb  (b) 950 in.−lb  (c) 3800 in.−lb  (d) 190,000 in.−lb

$${It}\:{takes}\:{a}\:{force}\:{of}\:\mathrm{19},\mathrm{000}\:{lb}\:{to} \\ $$$${compress}\:{a}\:{spring}\:{from}\:{its}\:{free} \\ $$$${height}\:{of}\:\mathrm{15}\:{in}\:{to}\:{its}\:{fully}\: \\ $$$${compressed}\:{height}\:{of}\:\mathrm{10}\:{in}.\:{How} \\ $$$${much}\:\:{work}\:{does}\:{it}\:{take}\:{to}\: \\ $$$${compress}\:{the}\:{spring}\:{the}\:{first}\:{in}? \\ $$$$\left({a}\right)\:\mathrm{1900}\:{in}.−{lb} \\ $$$$\left({b}\right)\:\mathrm{950}\:{in}.−{lb} \\ $$$$\left({c}\right)\:\mathrm{3800}\:{in}.−{lb} \\ $$$$\left({d}\right)\:\mathrm{190},\mathrm{000}\:{in}.−{lb} \\ $$

Commented by mr W last updated on 08/Dec/20

W=(1/2)×1×((1/(10))×19000)=950 in.−lb  ⇒answer (b)

$${W}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\left(\frac{\mathrm{1}}{\mathrm{10}}×\mathrm{19000}\right)=\mathrm{950}\:{in}.−{lb} \\ $$$$\Rightarrow{answer}\:\left({b}\right) \\ $$

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