Certainly! Let's go through the problem step-by-step.
1. Identify the variables and the known values:
- Mass ([tex]\(m\)[/tex]) = 5 kg
- Acceleration ([tex]\(a\)[/tex]) = 4 m/s[tex]\(^2\)[/tex]
- Displacement ([tex]\(x\)[/tex]) = 2 m
2. Determine the force ([tex]\(F\)[/tex]) using Newton's Second Law:
According to Newton's Second Law of Motion, force is calculated as:
[tex]\[
F = m \times a
\][/tex]
Plugging in the values:
[tex]\[
F = 5 \, \text{kg} \times 4 \, \text{m/s}^2 = 20 \, \text{N}
\][/tex]
Therefore, the force [tex]\(F\)[/tex] is 20 Newtons.
3. Relate force ([tex]\(F\)[/tex]) to spring constant ([tex]\(k\)[/tex]) using Hooke's Law:
According to Hooke's Law, the force exerted by a spring is:
[tex]\[
F = -k \times x
\][/tex]
We need to solve for the spring constant [tex]\(k\)[/tex]. Rearranging the equation to solve for [tex]\(k\)[/tex]:
[tex]\[
k = -\frac{F}{x}
\][/tex]
Substitute the known values of [tex]\(F\)[/tex] and [tex]\(x\)[/tex]:
[tex]\[
k = -\frac{20 \, \text{N}}{2 \, \text{m}} = -10 \, \text{N/m}
\][/tex]
Therefore, the spring constant [tex]\(k\)[/tex] is [tex]\(-10 \, \text{N/m}\)[/tex].
Summary:
- The force [tex]\(F\)[/tex] is [tex]\(20 \, \text{N}\)[/tex].
- The spring constant [tex]\(k\)[/tex] is [tex]\(-10 \, \text{N/m}\)[/tex].
So, the values obtained are:
- Force: [tex]\(20 \, \text{N}\)[/tex]
- Spring Constant: [tex]\(-10 \, \text{N/m}\)[/tex]
