How to solve this. How long will it take the smaller block to make it to the right side of the larger block? Please give your answer in s .

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.

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