A spring is 17.0 cm long when it is lying on a table. Oneend is then attached to

Question

A spring is 17.0 cm long when it is lying on a table. Oneend is then attached to a hook and the other end is pulled by aforce that increases to 25.0 N, causing the spring to stretchto a length of 19.2 cm. a) How much work was required to stretch thespring from 17.0 cm to 19.2 cm? b)How long will the spring be if the25.0 N force is replaced by a 50.0 N force? A spring is 17.0 cm long when it is lying on a table. Oneend is then attached to a hook and the other end is pulled by aforce that increases to 25.0 N, causing the spring to stretchto a length of 19.2 cm. a) How much work was required to stretch thespring from 17.0 cm to 19.2 cm? b)How long will the spring be if the25.0 N force is replaced by a 50.0 N force? How much work was required to stretch thespring from 17.0 cm to 19.2 cm? b)How long will the spring be if the25.0 N force is replaced by a 50.0 N force?

Explanation / Answer

given: initial length=17 cm final length=19.2 cm then change in length becomes =2.2 cm= 0.022 m force constant k=F/change in length then k=25/0.022 approximatelyequals to 1136 N/m a) The work done by the force is stored as potential energy PE. So PE change of the spring equals the work done. PE=(1/2)kx^2 Initial PE is zero since the spring has no energy. Final PE is found by PEfinal= (1/2)*1136*(0.022)^2=0.275 Joule. b) if the force is doubled, the spring is streched by double of 0.022since k must be the same as in (1). Therefore it extends by 0.044mor 4.4 cm from 17 cm rest length and final length becomes17+4.4=21.4 cm

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