Respuesta :
Answer:
The answers to the question are
(a) 2.1 m/s
(b) 0.83 N
(c) 1.9 N
Explanation:
To solve the question, we list out the varibles
Length, l of string = 0.8 m
mass of rock, m = 0.12 kg
Angle with the verrticakl, θ = 45 °
a) To find the speed of the rock when the string passes through the vertical position we have
From the first law of thermodynamics
Potential energy = kinetic energy
m×g×l×(1-cosθ) = 1/2×m×v²
That is v² = 2×g×l×(1-cosθ)
= 2×9.81×0.8×(1-cos45) = 4.597
or v = √4.597 = 2.1 m/s
(b) The tension in the string when it makes an angle of 45∘ with the vertical is given by
For balance between Tension and mass of rock is gigen by
∑Forces = 0, T - m×g×cosθ = 0
or T = m×g×cosθ = 0.12×9.81×cos45 = 0.83 N
c) The tension in the string as it passes through the vertical
when passing through the vertical we have T - m×g = (m×v²)/r
or T = m×g + (m×v²)/r = mg(1+2(1-cosθ)) =0.981*0.12 (1+ 2(1-cos45)) =1.867 N
= 1.9 N
(A) the speed of the rock when the string passes the vertical position is 2.1 m/s
(B) the tension in the string when the angle is 45 degrees is 0.83 N
(B) the tension in the string when it passes the vertical position is 1.9 N
Circular motion:
(A)From the law of conservation of energy we can say that the potential energy is converted into kinetic energy as shown below:
[tex]mgl(1-cos\theta) = \frac{1}{2} mv^2\\\\v^2 = 2gl(1-cos\theta)\\\\v^2 = 2\times9.81\times0.8(1-cos45)\\\\v = 2.1 m/s[/tex]
(B) The tension in the string when it makes an angle of θ=°45 with the vertical is given by
T - mgcosθ = 0
T = mgcosθ
T = 0.12×9.81×cos45
T = 0.83 N
(C) The tension in the string as it passes through the vertical is given by:
[tex]T - mg = \frac{mv^2}{l}[/tex]
from (A) we have [tex]\frac{1}{2}mv^2=mgl(1-cos\theta)[/tex]
[tex]T = mg(1+2(1-cos\theta))\\\\T=0.981\times0.12 (1+ 2(1-cos45))\\\\T=1.9 N[/tex]
Learn more about circular motion:
https://brainly.com/question/2285236?referrer=searchResults
