Final answer:
The sphere's speed is approximately 2.9 m/s when it reaches the bottom of the ramp, as found by applying the conservation of energy principle and accounting for both translational and rotational kinetic energy.
Step-by-step explanation:
To find the sphere's speed when it reaches the lowest end of the board, we can use the principles of conservation of energy. Potential energy at the start will be converted into translational kinetic energy and rotational kinetic energy at the bottom.
The potential energy (PE) of the sphere at the height of 60 cm is given by PE = mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height. Since the sphere rolls without slipping, we must account for both translational kinetic energy (KE_trans = 1/2 m v^2, where v is the velocity of the sphere) and rotational kinetic energy (KE_rot = 1/2 I ω^2, where I is the moment of inertia of the sphere and ω is the angular velocity).
The moment of inertia for a solid sphere is I = 2/5 m r^2. Because the sphere rolls without slipping, v = rω. Thus, the total kinetic energy at the bottom is KE_total = KE_trans + KE_rot = 1/2 m v^2 + 1/2 (2/5 m r^2) (ω^2). We can find ω by rearranging v = rω to ω = v/r and substituting into our kinetic energy equation.
Conserving energy, PE = KE_total, so:
mgh = 1/2 m v^2 + 1/2 (2/5 m r^2) (v/r)^2
mgh = 1/2 m v^2 + 1/5 m v^2
mgh = 7/10 m v^2.
Solving for v gives us:
v^2 = (10/7) gh
v = √((10/7) gh)
v = √((10/7) (9.8 m/s^2) (0.60 m))
v = √(8.4) m/s.
The numerical solution yields v ≈ 2.9 m/s.