Answer:
1.8 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)
v = 5.47 m/s
The acceleration down the incline is constant, so given:
Δx = 4.8 m
v₀ = 0 m/s
v = 5.47 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (4.8 m) / (5.47 m/s + 0 m/s)
t = 1.76 s
Rounding to two significant figures, it takes 1.8 seconds.
The change in potential energy of an object is given by
where
m is the mass of the object
g is the gravitational acceleration
is the increase in altitude of the object
In our problem,
is the mass of the book,
and
is the increase in altitude of the book, so its variation of potential energy is
It’s going 3m/s. If we have 5 seconds to work with then we can find the acceleration by adding speed and how fast it going every second. So like the 7-10-13-16-19 so we go 3m/s faster ever second
First one is G second one is C.
