Question

At her maximum height a girl in a swing is 3m above the ground and at the lowest point she is 2m above the ground. Her maximum velocity is

• A. \(\sqrt{29.4}\) m/s
• B. \(\sqrt{9.8}\) m/s
• C. \(\sqrt{19.6}\) m/s
• D. 9.8 m/s

Hint:

In energy conservation problems, the change in kinetic energy is equal to the negative of the change in potential energy: \(\Delta KE = -\Delta PE\). Here, the loss in potential energy, \(mg(h_{top} - h_{bottom})\), is converted into a gain in kinetic energy, \(\frac{1}{2}mv_{max}^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the maximum velocity of a girl on a swing, given her heights at the highest and lowest points of her swing. The maximum velocity occurs at the lowest point.

Step 2: Key Formula or Approach:
We will use the principle of conservation of mechanical energy. The total mechanical energy (Kinetic Energy + Potential Energy) at the highest point is equal to the total mechanical energy at the lowest point, assuming no air resistance.
\[ KE_{top} + PE_{top} = KE_{bottom} + PE_{bottom} \]
\[ \frac{1}{2}mv_{top}^2 + mgh_{top} = \frac{1}{2}mv_{bottom}^2 + mgh_{bottom} \]

Step 3: Detailed Explanation:
Let's define the given values:
- Height at the top, \(h_{top} = 3\) m.
- Height at the bottom, \(h_{bottom} = 2\) m.
- At the maximum height (top), the swing momentarily stops, so \(v_{top} = 0\).
- At the lowest point (bottom), the velocity is maximum, so \(v_{bottom} = v_{max}\).
- We use the standard value for acceleration due to gravity, \(g = 9.8\) m/s\(^2\).
Substitute these into the conservation of energy equation:
\[ \frac{1}{2}m(0)^2 + mgh_{top} = \frac{1}{2}mv_{max}^2 + mgh_{bottom} \]
\[ 0 + mgh_{top} = \frac{1}{2}mv_{max}^2 + mgh_{bottom} \]
The mass \(m\) cancels from all terms:
\[ gh_{top} = \frac{1}{2}v_{max}^2 + gh_{bottom} \]
Rearrange to solve for \(v_{max}^2\):
\[ \frac{1}{2}v_{max}^2 = gh_{top} - gh_{bottom} = g(h_{top} - h_{bottom}) \]
\[ v_{max}^2 = 2g(h_{top} - h_{bottom}) \]
Substitute the numerical values:
\[ v_{max}^2 = 2(9.8)(3 - 2) = 2(9.8)(1) = 19.6 \]
\[ v_{max} = \sqrt{19.6} \text{ m/s} \]

Step 4: Final Answer:
Her maximum velocity is \(\sqrt{19.6}\) m/s.