Question

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are ăÿÿfrom the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Answer 1

We can use conservation of energy to solve this problem. At the initial position, the child has no kinetic energy and all her energy is potential energy due to her height above the lowest point of the swing. At the bottom of the swing, the child has no potential energy and all her energy is kinetic energy due to her speed.

(a) The potential energy of the child just as she is released can be calculated as:
PE = mgh
where m is the mass of the child, g is the acceleration due to gravity, and h is the height of the child above the lowest point of the swing. At the initial position, h = 2.20 m, so the potential energy is:
PE_initial = mgh = (25 kg)(9.81 m/s^2)(2.20 m) = 544 J

At the bottom of the swing, h = 0, so the potential energy is zero:
PE_bottom = 0 J

The potential energy at the initial position is greater than the potential energy at the bottom of the swing, since the child loses potential energy as she swings down.

(b) We can use conservation of energy to find the speed of the child at the bottom of the swing. At the initial position, all the energy is potential energy. At the bottom of the swing, all the energy is kinetic energy. Therefore, the potential energy at the initial position is equal to the kinetic energy at the bottom of the swing:
PE_initial = KE_bottom
mgh = (1/2)mv^2
where v is the speed of the child at the bottom of the swing. Solving for v, we get:
v = sqrt(2gh)
where sqrt means square root. Substituting the values, we get:
v = sqrt(2(9.81 m/s^2)(2.20 m)) = 6.26 m/s

Therefore, the child will be moving at a speed of 6.26 m/s at the bottom of the swing.

(c) The work done by the tension in the ropes as the child swings from the initial position to the bottom can be found as the change in the total mechanical energy of the child:
W = ΔE = KE_bottom - PE_initial
Substituting the values, we get:
W = (1/2)mv^2 - mgh
W = (1/2)(25 kg)(6.26 m/s)^2 - (25 kg)(9.81 m/s^2)(2

Answer 2

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Step-by-step explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

=

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J