To find the force $F$ needed to maintain constant speed, we analyze the forces acting on the sled while it is being pushed. The goal is to overcome friction by balancing it with the horizontal component of the applied force.

Given Data

• Mass $m = 31 \, \text{kg}$
• Coefficient of kinetic friction $\mu_k = 0.16$
• Angle $\theta = 25^\circ$
• Constant speed (implying zero acceleration)

Step 1: Calculate the Gravitational Force

The gravitational force $F_g$ acting on the sled is: $F_g = m \cdot g$ where $g = 9.81 \, \text{m/s}^2$.

$F_g = 31 \cdot 9.81 = 303.11 \, \text{N}$

Step 2: Decompose the Applied Force $F$

When pushing at an angle of $\theta = 25^\circ$ below the horizontal, the applied force $F$ has two components:

• Horizontal component: $F \cos \theta$
• Vertical component (downward): $F \sin \theta$

Step 3: Calculate the Normal Force

The normal force $N$ is affected by both the gravitational force and the vertical component of the applied force. Since $F \sin \theta$ acts downward, it adds to the weight:

$N = F_g + F \sin \theta$

Step 4: Calculate the Frictional Force

The kinetic frictional force $f_k$ is given by: $f_k = \mu_k \cdot N = \mu_k \cdot (F_g + F \sin \theta)$

Step 5: Set up the Equation for Constant Speed

For constant speed, the horizontal component of the applied force must balance the frictional force:

$F \cos \theta = f_k$

Substitute $f_k = \mu_k \cdot (F_g + F \sin \theta)$:

$F \cos \theta = \mu_k \cdot (F_g + F \sin \theta)$

Step 6: Solve for $F$

Rearrange the equation to isolate $F$:

$F \cos \theta - \mu_k F \sin \theta = \mu_k F_g$

Factor $F$ from the left side:

$F (\cos \theta - \mu_k \sin \theta) = \mu_k F_g$

Finally, solve for $F$:

$F = \frac{\mu_k F_g}{\cos \theta - \mu_k \sin \theta}$

Step 7: Substitute Values

Using $\mu_k = 0.16$, $F_g = 303.11 \, \text{N}$, and $\theta = 25^\circ$:

$F = \frac{0.16 \cdot 303.11}{\cos 25^\circ - 0.16 \sin 25^\circ}$

Calculating each term:

1. $\cos 25^\circ \approx 0.9063$
2. $\sin 25^\circ \approx 0.4226$

Substitute:

$F = \frac{0.16 \cdot 303.11}{0.9063 - (0.16 \cdot 0.4226)}$ $F = \frac{48.4976}{0.9063 - 0.0676}$ $F = \frac{48.4976}{0.8387} \approx 57.8 \, \text{N}$

Answer

The magnitude of the force needed to maintain the sled's motion at a constant speed is: $\boxed{57.8 \, \text{N}}$

Would you like a detailed breakdown of any specific step?

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Follow-up Questions:

1. What happens to the required force if the angle increases to 30°?
2. How would the answer change if the coefficient of kinetic friction were 0.2?
3. What effect would a different gravitational constant (like on another planet) have on the force calculation?
4. How does pulling at an angle above the horizontal (part b) affect the normal force?
5. How would this problem differ if the motion were on an inclined plane?

Tip: Always resolve forces into components parallel and perpendicular to the motion’s direction for clearer calculations.