Final answer:
The linear speed of a solid cylinder rolling down a 2.0 m high incline when it reaches the bottom is approximately 5.29 m/s, corresponding to choice A.
Step-by-step explanation:
The question involves finding the linear speed of a solid cylinder as it rolls down an inclined plane. Using conservation of energy and knowing that the cylinder starts from rest and the height of the incline is 2.0 m, we can calculate the final velocity. The potential energy at the top (mgh) is converted into translational kinetic energy (½mv²) and rotational kinetic energy (½Iω²). For a solid cylinder, the moment of inertia (I) is ½MR², and ω (angular velocity) is related to v (linear velocity) by ω = v/R. Considering energy conservation, we have:
mgh = ½mv² + ½(½MR²)(v/R)²
Solving for v gives:
v = (4gh/3)¹/²
For h = 2.0 m and g = 9.8 m/s² (acceleration due to gravity), we calculate:
v = (4×9.8×2.0/3)¹/² m/s ≈ 5.29 m/s
Hence, the linear speed at the bottom of the incline is approximately 5.29 m/s, which corresponds to choice A.