Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation
[tex]\Delta E_k=W=W_g-W_p[/tex]
where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:
[tex]Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J[/tex]
the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:
[tex]\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2[/tex]
where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:
[tex]v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}[/tex]
the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy
[tex]\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J[/tex]
hence, the change in Er is about 1.52J times the initial rotational energy
This question involves the concepts of the law of conservation of energy and kinetic energy.
a) The increase in translational kinetic energy is "0.139 J".
b) The new speed of yo-yo is "3.6 m/s".
c) The increase in rotational kinetic energy of yo-yo is "0.54 times" the initial rotational kinetic energy.
a)
The increase in translational kinetic energy can be found using the law of conservation of energy in the vertical direction:
[tex]Change\ in\ Transational\ Kinetic\ Energy = Work\ Done\ by\ Gravity-Work\ done\ by\ hand\\\\\Delta K.E=mgh-fd[/tex]where,
m = mass of yo-yo = 0.062 kg
g = acceleration due to gravity = 9.81 m/s²
h = height lost by yo-yo due to gravity = 0.32 m
f = force applied by hand = 0.35 N
d = distance moved by hand = 0.16 m
Therefore,
[tex]\Delta K.E = (0.062\ kg)(9.81\ m/s^2)(0.32\ m)-(0.35\ N)(0.16\ m)\\\Delta K.E = 0.195\ J - 0.056\ J\\[/tex]
ΔK.E = 0.139 J
b)
using the change in kinetic energy:
[tex]\Delta K.E = \frac{1}{2}(v_f^2-v_i^2)\\\\[/tex]
where,
vf = final speed = ?
vi = initial speed = 2.9 m/s
Therefore,
[tex]0.139\ J = \frac{1}{2}(0.062\ kg)(v_f^2-(2.9\ m/s)^2)\\\\(0.031\ kg)v_f^2=0.139\ J+0.261\ J\\v_f = \sqrt{\frac{0.4\ J}{0.031\ kg}}\\\\v_f = 3.6\ m/s[/tex]
c)
The change in rotational kinetic energy can be found as follows:
[tex]\frac{K.E_{rf}}{K.E_{if}}=\frac{\frac{1}{2}I\Omega_f^2}{\frac{1}{2}I\Omega_i^2}\\\\\frac{K.E_{rf}}{K.E_{if}}=\frac{\Omega_f^2}{\Omega_i^2}\\\\[/tex]
where,
[tex]K.E_{rf}[/tex] = final rotational kinetic energy
[tex]K.E_{if}[/tex] = initial rotational kinetic energy
[tex]\Omega_f[/tex] = final angular velocity = [tex]\frac{v_f}{r}=\frac{3.6\ m/s}{r}[/tex]
[tex]\Omega_i[/tex] = initial angular velocity = [tex]\frac{v_i}{r}=\frac{2.9\ m/s}{r}[/tex]
Therefore,
[tex]\frac{K.E_{rf}}{K.E_{if}}=\frac{(\frac{3.6\ m/s}{r})^2}{(\frac{2.9\ m/s}{r})^2}\\\\\frac{K.E_{rf}}{K.E_{if}}=\frac{(3.6\ m/s)^2}{(2.9\ m/s)^2}\\\\K.E_{rf}= 1.54\ K.E_{if}[/tex]
Now, the change in rotational kinetic energy is given as follows:
[tex]\Delta K.E_r=K.E_{rf}-K.E_{if}\\\\\Delta K.E_r=1.54\ K.E_{if}-K.E_{if}\\\\\Delta K.E_r=0.54\ K.E_{if}[/tex]
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
