(a) Free-Body Diagram

In the free-body diagram, the following forces act on the sled:

1. Gravitational force (mg): Acts vertically downward.
2. Normal force (N): Perpendicular to the slope.
3. Frictional force (f_k = μ_k N): Opposes the motion along the slope.
4. Component of gravitational force along the slope (mg sin α): Drives the sled down the slope.
5. Component of gravitational force perpendicular to the slope (mg cos α): Balanced by the normal force.

The diagram includes:

• Slope at an angle $\alpha$ with the horizontal.
• Forces labeled as $N$, $f_k$, $mg$, $mg \sin \alpha$, and $mg \cos \alpha$.

(b) Derivation of the acceleration

The net force $F_{\text{net}}$ acting along the slope is the difference between the downslope gravitational force and the frictional force:

$F_{\text{net}} = mg \sin \alpha - f_k$

Since $f_k = \mu_k N$ and $N = mg \cos \alpha$, we have:

$f_k = \mu_k (mg \cos \alpha)$

Thus, the net force becomes:

$F_{\text{net}} = mg \sin \alpha - \mu_k mg \cos \alpha$

Using Newton's second law $F = ma$, the acceleration $a$ is:

$a = \frac{F_{\text{net}}}{m} = g \sin \alpha - \mu_k g \cos \alpha$

Final expression for acceleration:

$a = g (\sin \alpha - \mu_k \cos \alpha)$

(c) Condition for constant speed ($a = 0$)

For constant speed, the acceleration $a = 0$. Using the derived expression for $a$:

$0 = g (\sin \alpha - \mu_k \cos \alpha)$

Simplify:

$\sin \alpha = \mu_k \cos \alpha$

Divide both sides by $\cos \alpha$ (valid for $\cos \alpha \neq 0$):

$\tan \alpha = \mu_k$

Thus, the sled moves at constant speed when:

$\alpha = \arctan(\mu_k)$

(d) Effect of mass $m$ and sign of acceleration

1. Does $m$ affect acceleration? The acceleration expression $a = g (\sin \alpha - \mu_k \cos \alpha)$ does not depend on $m$. This is because the mass $m$ cancels out when deriving the net force and acceleration.

2. Sign of acceleration and its dependence on $\alpha$:

   • For $\sin \alpha > \mu_k \cos \alpha$, $a > 0$, meaning the sled accelerates downslope.
   • For $\sin \alpha = \mu_k \cos \alpha$, $a = 0$, meaning the sled moves at constant speed.
   • For $\sin \alpha < \mu_k \cos \alpha$, $a < 0$, meaning the sled decelerates (or moves upslope if initially pushed upward).

Summary

• (a) Free-body diagram shows all forces acting on the sled.
• (b) Acceleration: $a = g (\sin \alpha - \mu_k \cos \alpha)$.
• (c) Constant speed occurs at $\alpha = \arctan(\mu_k)$.
• (d) $m$ does not affect $a$, and the sign of $a$ depends on the relative magnitudes of $\sin \alpha$ and $\mu_k \cos \alpha$.

Would you like detailed sketches or further clarification on any part?

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Related Questions:

1. How does the normal force change with varying $\alpha$?
2. What happens to acceleration if $\mu_k = 0$?
3. Can this system be modeled on a frictionless slope? How does the solution change?
4. How is the angle of constant speed affected if $\mu_k$ increases?
5. Can you derive the work-energy relationship for this sled system?

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Tip:

Always analyze forces perpendicular and parallel to the slope independently to avoid confusion in solving inclined-plane problems!