Question

1. 19. a spring with k = 100 n/m is stretched by 0.1 m. what is the force? hint: use f = k x. 22. 20. what is the spring constant if a spring stores 1.25 j of energy when stretched 0.25 m? hint: use pe = 1/2 k * x².

Explanation:

Step1: Solve for force in first - question

Use Hooke's law formula $F = kx$. Given $k = 100\ N/m$ and $x=0.1\ m$.
$F=100\times0.1$

Step2: Calculate the value of force

$F = 10\ N$

Step3: Solve for spring - constant in second - question

Use the elastic - potential energy formula $PE=\frac{1}{2}kx^{2}$. Given $PE = 1.25\ J$ and $x = 0.25\ m$. First, re - arrange the formula for $k$: $k=\frac{2PE}{x^{2}}$.
$k=\frac{2\times1.25}{(0.25)^{2}}$

Step4: Calculate the value of spring - constant

$k=\frac{2.5}{0.0625}=40\ N/m$

Answer:

1. The force is $10\ N$.
2. The spring constant is $40\ N/m$.