# Answer to Question #27115 in Calculus for jan

Question #27115

find the work done in stretching a spring of natural length 8cm from 10cm to 13cm. assume a force of 6N is needed to hold it a length of 11cm.

Expert's answer

find the work done in stretching a spring of natural length 8cmfrom 10cm to 13cm. assume a force of 6N is needed to hold it a length of 11cm.

Hooke's law:F = k*Δl

k - stiffness

Δl = l - l0,

l0 - naturallength, l - length after stretching.

In our case: 6 H =k*(11cm - 8cm) => k = 6H/3cm = 2H/cm

Energy of deformationequals:

E = k*(l-l0)^2/2

The law of conservationof energy:

E1 + A = E2

A - the work donein stretching a spring

E1 = k*(l1 - l0)^2/2, E2 = k*(l2 - l0)^2/2

l1 = 10 cm, l2 = 13 cm

Finally:

A = k*(l2 - l0)^2/2 - k*(l1 - l0)^2/2 = 2 H/cm*(13cm - 8cm)^2/2 - 2 H/cm*(10cm - 8cm)^2/2

= 5^2 H*cm - 2^2 H*cm = 21 H*cm = 0.21 H*m = 0.21 J

Answer: the workdone in stretching a spring equals 0.21 J

Hooke's law:F = k*Δl

k - stiffness

Δl = l - l0,

l0 - naturallength, l - length after stretching.

In our case: 6 H =k*(11cm - 8cm) => k = 6H/3cm = 2H/cm

Energy of deformationequals:

E = k*(l-l0)^2/2

The law of conservationof energy:

E1 + A = E2

A - the work donein stretching a spring

E1 = k*(l1 - l0)^2/2, E2 = k*(l2 - l0)^2/2

l1 = 10 cm, l2 = 13 cm

Finally:

A = k*(l2 - l0)^2/2 - k*(l1 - l0)^2/2 = 2 H/cm*(13cm - 8cm)^2/2 - 2 H/cm*(10cm - 8cm)^2/2

= 5^2 H*cm - 2^2 H*cm = 21 H*cm = 0.21 H*m = 0.21 J

Answer: the workdone in stretching a spring equals 0.21 J

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