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Chapter 5: Problem 30

A mother pulls her daughter, whose mass is \(20.0 \mathrm{~kg}\) and who is sitting on a swing with ropes of length \(3.50 \mathrm{~m}\), backward until the ropes make an angle of \(35.0^{\circ}\) with respect to the vertical. She then releases her daughter from rest. What is the speed of the daughter when the ropes make an angle of \(15.0^{\circ}\) with respect to the vertical?

Short Answer

Expert verified
Answer: To find the speed of the child when the ropes make an angle of 15.0° with respect to the vertical, we can apply the conservation of mechanical energy. After calculating the initial and final heights using the given angles, and substituting the known values for mass, gravitational acceleration, and height differences, we can find the final speed of the child. To calculate the speed, use the following formula: $$ v = \sqrt{2(9.81\mathrm{m/s^2})(h_i - h_f)} $$ Plug in the calculated initial and final heights and solve for the final speed, v, to get the answer to the problem.

Step by step solution

01

Identify known and unknown variables

We know the following variables: - Mass of the child, \(m = 20.0\mathrm{~kg}\) - Length of the swing ropes, \(L = 3.50\mathrm{~m}\) - Initial angle with respect to the vertical, \(\theta_1 = 35.0^{\circ}\) - Final angle required, \(\theta_2 = 15.0^{\circ}\) We need to find the final speed, \(v\).
02

Calculate the initial and final heights

We will first find the initial and final heights of the child with respect to the lowest point of the swing. To do this, we use the following relation for the heights, considering the length of the ropes and the respective angles: $$ h_i = L - L\cos{\theta_1} $$ $$ h_f = L - L\cos{\theta_2} $$ where \(h_i\) is the initial height and \(h_f\) is the final height.
03

Apply conservation of mechanical energy

Since only gravitational force is acting on the system, mechanical energy is conserved. Therefore, we can write the following equation using the potential energy and kinetic energy of the child at the initial and final instants (\(U\) for potential energy, \(K\) for kinetic energy): $$ K_i + U_i = K_f + U_f $$ Initially, the child is at rest, so the initial kinetic energy is zero. Thus, the equation becomes: $$ U_i = K_f + U_f $$ Potential energy is calculated using the mass, gravitational constant (\(g = 9.81\mathrm{m/s^2}\)) and heights: $$ mgh_i = \frac{1}{2}mv^2 + mgh_f $$ Now, we can solve the equation for the final speed, \(v\).
04

Calculate the final speed

We can rearrange the equation from step 3 to solve for \(v\): $$ v = \sqrt{2g(h_i - h_f)} $$ Using the values calculated in step 2 for the initial and final heights, and substituting the known values for mass and gravitational acceleration, we can find the final speed, \(v\), when the ropes make an angle of \(15.0^{\circ}\) with respect to the vertical.
05

Find and report the final value

Plug in the known values and calculate the final speed of the daughter: $$ v = \sqrt{2(9.81\mathrm{m/s^2})(h_i - h_f)} $$ Once the final speed is calculated, it can be reported as the answer to the problem.

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Most popular questions from this chapter

A car of mass \(m=1250 \mathrm{~kg}\) is traveling at a speed of \(v_{0}=105 \mathrm{~km} / \mathrm{h}(29.2 \mathrm{~m} / \mathrm{s}) .\) Calculate the work that must be done by the brakes to completely stop the car.

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