Question

• - f is the force applied (in newtons)
• - k is the spring constant (in n/m)
• - x is the extension or compression (in meters)

1. 2. potential energy in a spring: pe = 1/2 k x²

• - pe is the potential energy stored (in joules)

questions:

1. 1. a spring stretches 0.2 m when a force of 4 n is applied. what is the spring constant?

• hint: use f = k * x and solve for k.

1. 2. if a spring has a spring constant of 50 n/m and is compressed by 0.1 m, what is the force?

• hint: use f = k * x.

1. 3. a spring is stretched by 0.3 m and stores 2.25 j of energy. what is the spring constant?

• hint: use pe = 1/2 k x² and solve for k.

1. 4. what is the potential energy stored in a spring with k = 100 n/m and x = 0.2 m?

• hint: use pe = 1/2 k x².

1. 5. a spring requires 10 n to stretch 0.5 m. what is the spring constant?

• hint: use f = k * x.

1. 6. how much energy is stored in a spring with k = 30 n/m and x = 0.4 m?

• hint: use pe = 1/2 k x².

Explanation:

Step1: Solve for spring - constant in first question

Given $F = kx$, we can solve for $k$ by rearranging the formula to $k=\frac{F}{x}$. Substituting $F = 4N$ and $x = 0.2m$, we get $k=\frac{4}{0.2}$.
$k = 20N/m$

Step2: Solve for force in second question

Using the formula $F=kx$, with $k = 50N/m$ and $x = 0.1m$, we have $F=50\times0.1$.
$F = 5N$

Step3: Solve for spring - constant in third question

Given $PE=\frac{1}{2}kx^{2}$, we can solve for $k$ by rearranging the formula to $k=\frac{2PE}{x^{2}}$. Substituting $PE = 2.25J$ and $x = 0.3m$, we get $k=\frac{2\times2.25}{0.3^{2}}$.
$k=\frac{4.5}{0.09}=50N/m$

Step4: Solve for potential energy in fourth question

Using the formula $PE=\frac{1}{2}kx^{2}$, with $k = 100N/m$ and $x = 0.2m$, we have $PE=\frac{1}{2}\times100\times0.2^{2}$.
$PE = 2J$

Step5: Solve for spring - constant in fifth question

Given $F = kx$, we solve for $k$ by $k=\frac{F}{x}$. Substituting $F = 10N$ and $x = 0.5m$, we get $k=\frac{10}{0.5}$.
$k = 20N/m$

Step6: Solve for potential energy in sixth question

Using the formula $PE=\frac{1}{2}kx^{2}$, with $k = 30N/m$ and $x = 0.4m$, we have $PE=\frac{1}{2}\times30\times0.4^{2}$.
$PE=\frac{1}{2}\times30\times0.16 = 2.4J$

Answer:

1. $k = 20N/m$
2. $F = 5N$
3. $k = 50N/m$
4. $PE = 2J$
5. $k = 20N/m$
6. $PE = 2.4J$