Physics, 04.12.2019 10:31, joeylozier15

A25.0-kg child plays on a swing having support ropes that are 2.20 m long. her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) what is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) how fast will she be moving at the bottom? (c) how much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

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A) Her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion is; P.E = 138.61 J

B) The speed she will be moving with at the bottom is; v = 3.33 m/s

C) The amount of work that the tension in the ropes do as she swings from the initial position to the bottom of the motion is; W = 0 J

We are given;

Mass of child; m = 25 kg

Length of rope; L = 2.2 m

Angle of of pull; θ = 42°

A) Since the length of the rope is 2.2 m and her brother pulls her back until the ropes are 42.0° from the vertical. Then, the difference in height between the initial and final position is;

Δh = L - Lcos θ

Δh = 2.2 - 2.2 cos 42°

Δh = 0.56518 m

Formula for potential energy between two heights is;

P.E = mgΔh

P.E = 25 × 9.81 × 0.56518

P.E = 138.61 J

B) To find how fast will she be moving at the bottom, we will make use of conservation of energy where;

P.E = K.E

Formula for Kinetic energy (K.E) is;

K.E = ¹/₂mv²

where v is the speed at the bottom.

Thus;

¹/₂mv² = mgΔh

making v the subject gives;

v = √(2gΔh)

v = √(2 × 9.81 × 0.56518)

v = 3.33 m/s

C) Formula for work done in this case by the rope is;

W = T(d cos θ)

Where T is tension in rope and W is workdone

Now, when swinging the tension in rope would be perpendicular to the distance covered by the child and as such the angle is θ = 90°

From trigonometric functions, cos 90° = 0

Thus; W = 0 J

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answered: TheIncognateo

(a) 139.7 J

The potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

$U=mg\Delta h$

where

m = 25.0 kg is the mass of the child

g = 9.8 m/s^2

$\Delta h$ is the difference in height between the initial position and the bottom position

We are told that the rope is L = 2.20 m long and inclined at 42.0° from the vertical: therefore, $\Delta h$ is given by

$\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m$

So, her potential energy is

U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J

(b) 3.3 m/s

At the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

$U=K=\frac{1}{2}mv^2$

where

m = 25.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s$

The work done by the tension in the rope is given by:

$W=Td cos \theta$

where

T is the tension

d is the displacement of the child

$\theta$ is the angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. Therefore, $\theta=90^{\circ}$ and $cos \theta=0$ , so the work done is zero.