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andreev551 [17]
3 years ago
14

A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se

conds after starting from rest?
312 J


237 J


156 J


106 J
Physics
1 answer:
grigory [225]3 years ago
6 0
First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)
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Answer:

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\Delta E_k=W=W_g-W_p

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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:

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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:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

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}

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c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\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

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Answer:

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Explanation:

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