Question

Suppose a force of 30 N is required to stretch and hold a spring 0.2 m from its equilibrium position.

a. Assuming the spring obeys​ Hooke's law, find the spring constant k.
b. How much work is required to compress the spring 0.1 m from its equilibrium​ position?
c. How much work is required to stretch the spring 0.4 m from its equilibrium​ position?
d. How much additional work is required to stretch the spring 0.2 m if it has already been stretched 0.2 m from its​ equilibrium?

Answer 1

a. k=150N/m, b.W=0.75J,c. W=12J, d.W=9J

Step-by-step explanation:

a. The Hooke's Law states
`F=kx`, where
`F` is the force,
`k` the spring constant and
`x` the displacement, Knowing the force and the displacement you can find
`k`:

`F=kx`

`k=\frac{F}x=(30)/(0.2)=150N/m`

b. The work done by a force
`F(s)` that moves along a displacement
`s` is:

`W=\int\limits^(x_2)_(x_1) {F(s)} \, ds`

Then
`F(s)=ks` (Hooke's Law)

`W=\int\limits^(x_2)_(x_1) {F(s)} \, ds=\int\limits^(x_2)_(x_1) {ks} \, ds=(1)/(2) ks^2|\limits^(x_2)_(x_1)=(1)/(2) k(x_2^2-x_1^2)`

Work needed to go from
`x_1=0` to
`x_2=0.1`

`W=(1)/(2) k(x_2^2-x_1^2)=(1)/(2) 150(0.1^2-0^2)=75(0.01)=0.75 J`

c. Work needed to go from
`x_1=0` to
`x_2=0.4`

`W=(1)/(2) k(x_2^2-x_1^2)=(1)/(2) 150(0.4^2-0^2)=75(0.16)=12 J`

d. Work needed to go from
`x_1=0.2` to
`x_2=0.4`

`W=(1)/(2) k(x_2^2-x_1^2)=(1)/(2) 150(0.4^2-0.2^2)=75(0.16-0.04)=75(0.12)=9 J`