Let’s calculate the quantities step by step.

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1. Acceleration of the block

We know:

• Initial velocity $u = 0$
• Distance $s = 2.00 \, \mathrm{m}$
• Time $t = 1.75 \, \mathrm{s}$

Using the kinematic equation:

$s = ut + \frac{1}{2} a t^2$

Substituting the known values:

$2.00 = 0 \cdot 1.75 + \frac{1}{2} a (1.75)^2$

$2.00 = \frac{1}{2} a \cdot 3.0625$

$a = \frac{2.00 \cdot 2}{3.0625} = 1.306 \, \mathrm{m/s^2}$

The acceleration of the block is:

$\boxed{1.31 \, \mathrm{m/s^2}}$

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2. Coefficient of friction

On an incline, the net force is given by:

$F_{\text{net}} = m a = m g \sin \theta - F_{\text{friction}}$

The frictional force is related to the normal force and the coefficient of friction $\mu$:

$F_{\text{friction}} = \mu F_{\text{normal}} = \mu m g \cos \theta$

Rewriting the net force equation:

$m a = m g \sin \theta - \mu m g \cos \theta$

Dividing through by $m g$:

$\frac{a}{g} = \sin \theta - \mu \cos \theta$

Substituting known values:

• $a = 1.31 \, \mathrm{m/s^2}$
• $g = 9.8 \, \mathrm{m/s^2}$
• $\theta = 26.5^\circ$

$\frac{1.31}{9.8} = \sin(26.5^\circ) - \mu \cos(26.5^\circ)$

$0.1337 = 0.4462 - \mu (0.8949)$

$\mu (0.8949) = 0.4462 - 0.1337$

$\mu = \frac{0.3125}{0.8949} = 0.349$

The coefficient of friction is:

$\boxed{0.349}$

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3. Frictional force acting on the block

The frictional force is given by:

$F_{\text{friction}} = \mu m g \cos \theta$

Substituting known values:

• $\mu = 0.349$
• $m = 3.00 \, \mathrm{kg}$
• $g = 9.8 \, \mathrm{m/s^2}$
• $\cos(26.5^\circ) = 0.8949$

$F_{\text{friction}} = 0.349 \cdot 3.00 \cdot 9.8 \cdot 0.8949$

$F_{\text{friction}} = 9.18 \, \mathrm{N}$

The frictional force acting on the block is:

$\boxed{9.18 \, \mathrm{N}}$

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4. Speed of the block after sliding 2.00 m

Using the kinematic equation:

$v^2 = u^2 + 2 a s$

Substituting known values:

• $u = 0 \, \mathrm{m/s}$
• $a = 1.31 \, \mathrm{m/s^2}$
• $s = 2.00 \, \mathrm{m}$

$v^2 = 0 + 2 (1.31)(2.00)$

$v^2 = 5.24$

$v = \sqrt{5.24} = 2.29 \, \mathrm{m/s}$

The speed of the block after sliding 2.00 m is:

$\boxed{2.29 \, \mathrm{m/s}}$

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Final Answers:

1. Acceleration: $\boxed{1.31 \, \mathrm{m/s^2}}$
2. Coefficient of friction: $\boxed{0.349}$
3. Frictional force: $\boxed{9.18 \, \mathrm{N}}$
4. Speed after sliding 2.00 m: $\boxed{2.29 \, \mathrm{m/s}}$

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Would you like further clarifications or details?

Related Questions:

1. How does the angle of incline affect the coefficient of friction?
2. What happens to the frictional force if the block's mass changes?
3. How is the acceleration influenced by increasing the coefficient of friction?
4. Can the coefficient of friction exceed 1? Why or why not?
5. How would the speed change if the incline were frictionless?

Tip:

When working with inclined planes, always resolve forces parallel and perpendicular to the incline for a clear