Question

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

Answer 1

Answer:

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

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 $k$:

$F=kx$

$k=\frac{F}x=\frac{30}{0.2}=150N/m$

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)

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

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

$W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.1^2-0^2)=75(0.01)=0.75 J$

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

$W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0^2)=75(0.16)=12 J$

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

$W=\frac{1}{2} k(x_2^2-x_1^2)=\frac{1}{2} 150(0.4^2-0.2^2)=75(0.16-0.04)=75(0.12)=9 J$