Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 52 cm.

Question

How much work is needed to stretch the spring from 39 cm to 47 cm?

How far beyond its natural length will a force of 35 N keep the spring stretched?

Answer 1

First consider the position of the terminal end of the spring is x= 0, if the spring is stretched from 34 cm to 52cm x will vary from 0 to 18cm, 18cm is the difference between 52 and 34

The needed work to stretch the spring from 39 cm to 47 cm is:

^0.18

W = ∫₀ F(x) dx

F( x) = k x , F(x) is the force in N and k is the spring constant in N/m

^0.18

W = ∫₀ k x dx

^0.18

2 = k( x² /2) Ι₀

2 = k/2 ( 0.18² )

k= 123.45 N/m

To stretch the spring from 39cm to 47cm x will change from (39-34) =5cm to (47-34)=13cm

^{0.13 0.13}

W = ∫_0.05 k x dx = k x² /2 Ι_0.05 =123. 45 /2 ( 0.13² - 0.05² ) = 0.88 J

F(x) = k x

35 = 123.45 x

x = 35/ 123.45 = 0.28 m =28cm

that is mean the spring will be stretched from its natural length 34cm to (34+28)=62cm

Answer 2

The work done equals the potential energy put into the spring.

Potential energy (U_S) for a spring is calculated from:

U_S = 1/2kx² where "x" is the distance the spring is stretched. Since I'd like to stay in SI units, I'll convert the stretch into meters: 52 cm = 0.52 m and 34 cm = 0.34 m so the "stretch" is 0.52 - 0.34 = 0.18 m

So 2 = 1/2k(0.18²)

4 = k(0.18²)

4 = 0.0324k

k = 123.45679 N/m

So the work required to stretch the spring from 34 to 39 (a total of 0.05 m) is:

U_S = 1/2kx²

U_S = 1/2(123.45679)(0.05²) = 0.1543 J

And the work required to stretch the spring from 34 to 47 (a total of 0.13 m) is:

U_S = 1/2kx²

U_S = 1/2(123.45679)(0.13²) = 1.0432 J

So the work required to go from 39 to 47 is 1.0432 - 0.1543 = 0.8889 J (I'll let you round to the appropriate sig fig)

The force of a stretched spring uses the equation F = -kx so:

-35 = -(123.45679)(x)

x = 0.28 m = 28 cm