Answer:
d) 14.1 m
e) 8.57 m/s
Explanation:
(a) There are three forces on the ball:
Normal force Fn pushing up,
Weight force Fg pulling down,
Kinetic friction Fk pushing left (resisting motion).
FBD is the correct free body diagram.
(b) Sum of forces in the y direction:
∑F = ma
Fn − Fg = 0
Fn = Fg
Sum of forces in the x direction:
∑F = ma
-Fk = ma
-Fn μk = ma
-mg μk = ma
a = -g μk
a = -(10 m/s²) (0.25)
a = -2.5 m/s²
(c) Sum of torques about the ball's center:
∑τ = Iα
Fk r = ⅖ mr² α
mg μk r = ⅖ mr² α
g μk = ⅖ r α
α = 5g μk / (2r)
α = 5 (10 m/s²) (0.25) / (2 × 0.2 m)
α = 31.25 rad/s²
(d) The velocity and angular velocity at time t is:
v = at + v₀
v = -2.5t + 12.0
ω = αt + ω₀
ω = 31.25t + 0
When the ball begins to roll without slipping, v = ωr.
-2.5t + 12.0 = (31.25t) (0.2)
-2.5t + 12.0 = 6.25t
12.0 = 8.75t
t = 1.37
The distance the ball moves is:
Δx = v₀ t + ½ at²
Δx = (12) (1.37) + ½ (-2.5) (1.37)²
Δx = 14.1 m
(e) v = at + v₀
v = (-2.5 m/s²) (1.37 s) + 12.0 m/s
v = 8.57 m/s
