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A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the hoop after it has rolled 3.00 m up as measured along the surface of the ramp

Respuesta :

Answer:

[tex]v_f = 4.22 m/s[/tex]

Explanation:

As we know by energy conservation

initial total energy = final total energy

so we have

[tex]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2 + mgL sin\theta[/tex]

so we have

[tex]v = R \omega[/tex]

[tex]I = mR^2[/tex]

[tex]mv_i^2 = mv_f^2 + mgL sin\theta[/tex]

[tex]5.50^2 = v_f^2 + (9.81)(3) sin25[/tex]

[tex]30.25 - 12.43 = v_f^2[/tex]

[tex]v_f = 4.22 m/s[/tex]