Answer:
3.56 m/s²
Explanation:
According to conservation of energy, the can's initial potential energy is converted to linear and rotational kinetic energy. After finding the can's final speed, a kinematic equation can be used to find its acceleration.
Initial potential energy equals the sum of the linear and rotational kinetic energies.
PE = KE + RE
mgh = ½ mv² + ½ Iω²
where m is mass, g is gravity, v is the linear speed, I is the moment of inertia, and ω is the angular speed.
For a solid cylinder, moment of inertia is I = ½ mr².
mgh = ½ mv² + ½ (½ mr²) ω²
mgh = ½ mv² + ¼ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ¼ mv²
mgh = ¾ mv²
gh = ¾ v²
v = √(⁴/₃ gh)
Plugging in values:
v = √(⁴/₃ (9.8 m/s²) (5.1 m sin 33°))
v = 6.02 m/s
Now we can use a kinematic equation to find acceleration. Given:
s = 5.1 m
u = 0 m/s
v = 6.02 m/s
Find: a
v² = u² + 2as
(6.02 m/s)² = (0 m/s)² + 2a (5.1 m)
a = 3.56 m/s²
