Question

k is the spring constant (in n/m)
x is the extension or compression (in meters)

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

• pe is the potential energy stored (in joules)

questions:

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

• hint: use f = k * x.

1. 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:

1.

Step1: Rearrange Hooke's law formula

Given $F = kx$, we can solve for $k$ as $k=\frac{F}{x}$.

Step2: Substitute values

Substitute $F = 4N$ and $x=0.2m$ into the formula. So $k=\frac{4}{0.2}=20N/m$.

Step1: Apply Hooke's law

Use the formula $F = kx$.

Step2: Substitute values

Substitute $k = 50N/m$ and $x = 0.1m$ into the formula. So $F=50\times0.1 = 5N$.

Step1: Rearrange potential - energy formula

Given $PE=\frac{1}{2}kx^{2}$, we can solve for $k$ as $k=\frac{2PE}{x^{2}}$.

Step2: Substitute values

Substitute $PE = 2.25J$ and $x = 0.3m$ into the formula. So $k=\frac{2\times2.25}{0.3^{2}}=\frac{4.5}{0.09}=50N/m$.

Answer:

$20N/m$

2.