- Acceleration: 1.3 m/s².
- Net Force: 80 N (2 sig. fig.).
- Weight: 5.9 × 10² N = 590 N.
Explanation
Apply the equation for objects that move in circles:
[tex]a = \dfrac{v^2}{r}[/tex],
where
- [tex]a[/tex] is the acceleration of the object,
- [tex]v[/tex] is the tangential speed of the object, and
- [tex]r[/tex] is the radius of the circular path.
For this monster:
- [tex]a[/tex] is to be found,
- [tex]v = 2\;m\cdot s^{-1}[/tex], and
- [tex]r = 3\;\text{m}[/tex].
[tex]\displaystyle a = \frac{({2\;\text{m}\cdot\text{s}^{-1}})^{2}}{3\;\text{m}} = \frac{2^2}{3}\;\text{m}\cdot\text{s}^{-2}= 1.3\;\text{m}\cdot\text{s}^{-2}[/tex].
By Newton's Second Law,
[tex]\Sigma F = m\cdot a[/tex],
where
- [tex]\Sigma F[/tex] is the net force on an object,
- [tex]m[/tex] is the mass of the object, and
- [tex]a[/tex] is the acceleration of the object.
- [tex]\Sigma F[/tex] is to be found,
- [tex]m = 60\;\text{kg}[/tex] for this monster, and
- [tex]a = 1.33333\;\text{m}\cdot\text{s}^{-2}[/tex] from previous calculations.
[tex]\Sigma F = m\cdot a = 60\;\text{kg} \times 1.33333\;\text{m}\cdot\text{s}^{-2} = 80\;\text{N}[/tex].
Weight of an object near the surface of the earth:
[tex]W = m\cdot g[/tex],
where
- [tex]m[/tex] is the mass of the object, and
- [tex]g[/tex] is the gravitational acceleration "constant" (a.k.a. gravitational field strength.) [tex]g \approx 9.81\;\text{N}\cdot\text{kg}^{-1}[/tex] near the surface of the earth.
[tex]W = 60\;\text{kg} \times 9.81\;\text{N}\cdot\text{kg}^{-1}=5.9\times 10^{2} \;\text{N}[/tex].
