A 100 kg solid spherical rock (I = 2/5 MR²) has a diameter of 50.0 cm. The rock is rolling down a hill with a velocity of 5.00 m/s. The total kinetic energy (angular + translational) of the rolling rock is A) 1,750 J. B) 2,000 J. C) 2,250 J. D) 2,670 J. E) 2,900 J.

Question

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The correct answer is A.The total kinetic energy of the rolling rock is the sum of its translational kinetic energy and its rotational kinetic energy. The translational kinetic energy is given by 1/2 mv^2, where m is the mass of the rock and v is its velocity. The rotational kinetic energy is given by 1/2 Iω^2, where I is the moment of inertia of the rock and ω is its angular velocity.The moment of inertia of a solid sphere is given by 2/5 MR^2, where M is the mass of the sphere and R is its radius. The angular velocity of the rock is given by v/R, where v is the velocity of the rock and R is its radius.Substituting these values into the equations for translational kinetic energy and rotational kinetic energy, we get:Translational kinetic energy = 1/2 (100 kg) (5.00 m/s)^2 = 1,250 JRotational kinetic energy = 1/2 (2/5 (100 kg) (0.25 m)^2) (5.00 m/s / 0.25 m)^2 = 500 JTotal kinetic energy = 1,250 J + 500 J = 1,750 JTherefore, the total kinetic energy of the rolling rock is 1,750 J.

A 100 Kg Solid Spherical Rock (I = 2/5 MR2)

A 100 Kg Solid Spherical Rock (I = 2/5 MR²)