Respuesta :
(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
[tex]U=mg\Delta h[/tex]
where
m = 25.0 kg is the mass of the child
g = 9.8 m/s^2
[tex]\Delta h[/tex] 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, [tex]\Delta h[/tex] is given by
[tex]\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m[/tex]
So, her potential energy is
[tex]U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J[/tex]
(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:
[tex]U=K=\frac{1}{2}mv^2[/tex]
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
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s[/tex]
(c) 0
The work done by the tension in the rope is given by:
[tex]W=Td cos \theta[/tex]
where
T is the tension
d is the displacement of the child
[tex]\theta[/tex] 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, [tex]\theta=90^{\circ}[/tex] and [tex]cos \theta=0[/tex], so the work done is zero.
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
Read more at; https://brainly.com/question/17151691
