Question

1. if a spring stretches 0.25 m under a force of 5 n, find the spring constant. hint: use f = k * x.
2. a spring with k = 60 n/m is compressed by 0.15 m. what is the force? hint: use f = k * x.
3. find the potential energy in a spring with k = 80 n/m and x = 0.3 m. hint: use pe = 1/2 k x².
4. a spring stores 1.6 j of energy when stretched 0.2 m. what is the spring constant? hint: use pe = 1/2 k x².
5. what force is needed to stretch a spring with k = 25 n/m by 0.4 m? hint: use f = k * x.
6. a spring is compressed by 0.2 m and stores 0.5 j of energy. find the spring constant. hint: use pe = 1/2 k x².
7. if a spring has a spring constant of 70 n/m, how much force is needed to stretch it 0.3 m? hint: use f = k * x.
8. calculate the energy stored in a spring with k = 45 n/m and x = 0.25 m. hint: use pe = 1/2 k x².
9. a spring stretches 0.1 m under a force of 2 n. what is the spring constant? hint: use f = k * x.
10. find the force needed to compress a spring with k = 90 n/m by 0.2 m. hint: use f = k * x.
11. a spring stores 3.6 j of energy when stretched 0.4 m. what is the spring constant? hint: use pe = 1/2 k x².
12. how much energy is stored in a spring with k = 20 n/m and x = 0.5 m? hint: use pe = 1/2 k x².

Explanation:

7.

Step1: Rearrange Hooke's law formula

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

Step2: Substitute values

We have $F = 5\ N$ and $x=0.25\ m$. So $k=\frac{5}{0.25}=20\ N/m$.

Step1: Apply Hooke's law

Using the formula $F = k\times x$, where $k = 60\ N/m$ and $x = 0.15\ m$.

Step2: Calculate the force

$F=60\times0.15 = 9\ N$.

Step1: Use potential - energy formula

The formula for the potential energy of a spring is $PE=\frac{1}{2}kx^{2}$, with $k = 80\ N/m$ and $x = 0.3\ m$.

Step2: Substitute values and calculate

$PE=\frac{1}{2}\times80\times(0.3)^{2}=40\times0.09 = 3.6\ J$.

Answer:

$20\ N/m$

8.