A 80- kg child sits in a swing suspended with 2.6- m-long ropes. The swing is held aside so that the ropes make an angle of 46 o with the vertical. Use conservation of energy to determine the speed the child will have at the bottom of the arc when she is let go.
(M/2)Vmax^2 = M g L(1 - cos46)
L(1- cos 46) is the maximum rise of the swing above the bottom position, where the velocity is Vmax
Vmax = sqrt[2 g L (1-cos46)]
does 1-cos46 represent L or is L part of that equation?
To determine the speed the child will have at the bottom of the swing, we can use the principle of conservation of energy. The total mechanical energy of the system remains constant throughout the motion. At the highest point, all of the energy is in the form of gravitational potential energy, and at the bottom point, it is in the form of kinetic energy.
First, let's find the gravitational potential energy at the highest point. The gravitational potential energy (Ug) is given by the formula:
Ug = mgh
where m is the mass (80 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height.
To find the vertical height, we can use trigonometry. The length of the ropes (L) and the angle with the vertical (θ) are given. The vertical height (h) can be calculated as:
h = L * sin(θ)
Substituting the given values, we have:
h = (2.6 m) * sin(46°)
Now, we can calculate the gravitational potential energy:
Ug = (80 kg) * (9.8 m/s^2) * (2.6 m) * sin(46°)
Next, we find the kinetic energy (K) at the bottom point. The kinetic energy is given by the formula:
K = 1/2 * mv^2
where m is the mass (80 kg) and v is the velocity at the bottom point.
Since the total mechanical energy is conserved, the gravitational potential energy at the highest point is equal to the kinetic energy at the bottom point:
Ug = K
Solving for v, we can rearrange the equation:
v = sqrt(2 * Ug / m)
Substituting the values we calculated for Ug and m:
v = sqrt(2 * [(80 kg) * (9.8 m/s^2) * (2.6 m) * sin(46°)] / (80 kg))
Simplifying further:
v = sqrt(2 * (9.8 m/s^2) * (2.6 m) * sin(46°))
v = sqrt(50.6) m/s
Therefore, the speed the child will have at the bottom of the arc when she is let go is approximately 7.11 m/s.
L is the rope length, and is part of the equation. I already told you what
L*(1- cos 46) is .