Answer:
8.1 s
Explanation:
Draw a free body diagram of the small block. There are four forces acting on the block:
Applied force F pushing to the right,
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing left.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces in the x direction:
∑F = ma
F − Nμ = ma
F − mgμ = ma
a = (F − mgμ) / m
Plug in values:
a = (12.2 N − (3.0 kg) (9.8 m/s²) (0.320)) / (3.0 kg)
a = 0.931 m/s²
Next, draw a free body diagram of the larger block. There are four forces:
Normal force N pushing down (equal and opposite),
Friction force Nμ pushing right (equal and opposite),
Weight force Mg pulling down,
and normal force N₂ pushing up.
Sum of forces in the x direction:
∑F = ma
Nμ = Ma
mgμ = Ma
a = mgμ / M
Plug in values:
a = (3.0 kg) (9.8 m/s²) (0.320) / (11.0 kg)
a = 0.855 m/s²
So the acceleration of the smaller block relative to the larger block is 0.931 m/s² − 0.855 m/s² = 0.0754 m/s².
Given:
Δx = 2.5 m
v₀ = 0 m/s
a = 0.0754 m/s²
Find: t
Δx = v₀ t + ½ at²
2.5 m = (0 m/s) t + ½ (0.0754 m/s²) t²
t = 8.14 s
Rounded to 2 significant figures, it takes 8.1 seconds.
