Tom is on a swing at a park. In order to swing higher, Sam applies a certain force to the swing. If the speed of the swing is 6 m/s at the lowest point, how high will the swing…
Tom is on a swing at a park. In order to swing higher, Sam applies a certain force to the swing. If the speed of the swing is 6 m/s at the lowest point, how high will the swing rise above the lowest point
To determine how high the swing will rise above the lowest point, we can use the principle of conservation of mechanical energy. At the lowest point of the swing, all the energy is kinetic, and as the swing rises, this kinetic energy is converted into potential energy.
The kinetic energy (KE) at the lowest point is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the swing (which will eventually cancel out), and \(v\) is the speed (6 m/s).
The potential energy (PE) at the highest point is given by the formula:
\[ PE = mgh \]
where \(g\) is the acceleration due to gravity (approximately 9.81 m/s²), and \(h\) is the height above the lowest point.
Setting the kinetic energy equal to the potential energy at the highest point gives us:
\[ \frac{1}{2} mv^2 = mgh \]
Since \(m\) is present in both sides of the equation, it cancels out, leading to:
\[ \frac{1}{2} v^2 = gh \]
Now we can solve for \(h\):
\[ h = \frac{v^2}{2g} \]
Substituting the given values:
\[ v = 6 \, \text{m/s}, \quad g = 9.81 \, \text{m/s}^2 \]
\[ h = \frac{(6 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]
\[ h = \frac{36}{19.62} \approx 1.837 \, \text{m} \]
Therefore, the swing will rise approximately 1.84 meters above the lowest point.
