Answer:
B ( 1.1 s )
Explanation:
Given that a thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal.
Let the height = h
Sin 30 = h / 3.1
h = 3.1 Sin 30
The P.E at the top = mgh
P.E = 9.8 × 3.1 Sin 30 × m
P.E = 15.19m
The K.E = P.E
K.E = 1/2mv^2
1/2 × V^2 × m = 15.19m
The mass m will cancel out.
V^2 = 30.38
V = 5.51 m/s
Using third equation of motion to find acceleration
V^2 = U^2 + 2as
Since it starts from rest, initial velocity u = 0
30.38 = 2 × 3.1 × a
30.38 = 6.2a
a = 30.38 / 6.2
a = 4.9 m/s^2
Using equation one of linear motion to calculate time t
V = U + at
5.51 = 4.9t
t = 5.51 / 4.9
t = 1.12 s
Therefore, it will take approximately 1.1 s to travel the first 3.1 m
The correct option is B
