Final answer:
The work required to stop a 40.0-kg solid cylinder rolling at a speed of 6.0 m/s is calculated using the work-energy principle, considering both translational and rotational kinetic energies, resulting in 720 J which corresponds to option (a).
Step-by-step explanation:
To determine how much work is required to stop a 40.0-kg solid cylinder rolling across a horizontal surface at a speed of 6.0 m/s, we can use the work-energy principle. The principle states that the work done is equal to the change in kinetic energy. In this case, we want to bring the cylinder to a stop, which means reducing its kinetic energy to zero.
For a rolling object, the total kinetic energy (KE) is the sum of its translational and rotational kinetic energy. The translational KE is given by ½mv², where m is the mass and v is the speed. The rotational KE is given by ½Iω², where I is the moment of inertia and ω is the angular velocity.
For a solid cylinder, I = ½mr² and ω = v/r.
Thus, the total KE is ½mv² + ½(½mr²)(v/r)² = ¾ mv².
Substituting the given values, KE = ¾(40.0 kg)(6.0 m/s)² = 720 J.
Therefore, the work required to stop the cylinder is 720 J.
The correct option is: (a) 720 J.