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Chapter 6: Problem 30

The maximum speed of a child on a swing is \(4.9 \mathrm{m} / \mathrm{s}\) The child's height above the ground is \(0.70 \mathrm{m}\) at the lowest point in his motion. How high above the ground is he at his highest point?

Short Answer

Expert verified
Answer: The child is about 1.93 meters above the ground at the highest point of their swing.

Step by step solution

01

Calculate the Kinetic Energy at the Lowest Point

To find the child's kinetic energy at the lowest point, we can use the following formula: \(KE = \frac{1}{2}mv^2\) where KE is the kinetic energy, m is the mass of the child, and v is the speed at the lowest point of the swing. We are given the speed (v = 4.9 m/s), but we do not have the mass to solve for KE directly. Make a mental note of this formula.
02

Calculate Potential Energy at the Lowest Point

At the lowest point of the swing, the child's height above the ground is 0.70 m. To find the potential energy at this height, we can use the following formula: \(PE_{low} = mgh\) where PE is the potential energy, m is the mass of the child, g is the acceleration due to gravity (9.81 \(\mathrm{m/s^{2}}\)), and h is the height above the ground. In this case, h = 0.70 meters.
03

Conservation of Energy

According to the law of conservation of energy, the total energy of the child (kinetic and potential) at the lowest point must be equal to the total energy of the child at the highest point. At the highest point, the child will momentarily have 0 velocity, so all the energy will be potential. Thus: \(KE_{low} + PE_{low} = PE_{high}\)
04

Substitute and Solve for the Height at the Highest Point.

We already know that \(KE_{low} = \frac{1}{2}mv^2\) and \(PE_{low} = mgh\), so we can substitute these expressions into our conservation of energy equation: \(\frac{1}{2}mv^2 + mgh = PE_{high}\) To resolve the height at the highest point, we can then isolate \(PE_{high}\): \(PE_{high} = \frac{1}{2}mv^2 + mgh\) Now, divide both sides by mg to get the height at the highest point (\(h_{high}\)): \(h_{high} = \frac{v^2}{2g} + h\) Plug in the given values (v = 4.9 m/s, h = 0.70 m, and g = 9.81 \(\mathrm{m/s^{2}}\)): \(h_{high} = \frac{(4.9)^2}{2(9.81)} + 0.70\)
05

Calculate the Result

Finally, evaluate the expression: \(h_{high} = \frac{24.01}{19.62} + 0.70 \approx 1.93 \mathrm{m}\) So, the child is about 1.93 meters above the ground at the highest point of their swing.

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