Final answer:
The total kinetic energy of a 1.2 kg solid disk rolling without slipping at 2.0 m/s is calculated as 3.6 J, which is rounded to 4 J.
Step-by-step explanation:
The question is related to the total kinetic energy of a rolling solid disk. In physics, the total kinetic energy of a rolling object consists of two parts: translational kinetic energy and rotational kinetic energy. For a disk rolling without slipping, the translational kinetic energy (KEt) can be calculated using the formula KEt = (1/2)mv2, where m is the mass and v is the velocity. The rotational kinetic energy (KEr) can be found using KEr = (1/2)Iω2, where I is the moment of inertia and ω is the angular velocity. For a solid disk, I = (1/2)mR2 and ω = v/R, where R is the radius. Hence, KEr = (1/2)(1/2)m(Rω)2 = (1/4)mv2. Adding both translational and rotational kinetic energies gives the total kinetic energy: KEtotal = KEt + KEr = (1/2)mv2 + (1/4)mv2 = (3/4)mv2.
For a 1.2 kg disk rolling at 2.0 m/s, the total kinetic energy is KEtotal = (3/4)(1.2 kg)(2.0 m/s)2 = (3/4)(1.2 kg)(4 m2/s2) = (3/4)(4.8 J) = 3.6 J. The correct answer, rounded to the nearest whole number as per the given options, would be 4 J, option a).