The girl and sled in (Figure 1) have a combined mass of 31 kg
. The coefficient of kinetic friction between the sled and the snow is 0.16.
Figure1 of 1
A person moves a girl on a sled along a horizontal snow surface. The figure consists of two parts labeled 'a' and b. In part 'a', the person pushes the girl with force F directed at angle theta below the horizontal. In part b, the person pulls on a rope tied to the sled with force F directed at angle theta above the horizontal.
Part A
If you push the sled, as in part (a) of the figure, at an angle of 25 ∘
 , what magnitude of force is needed to maintain the sled’s motion at a constant speed?
Express your answer with the appropriate units and use 3 significant figures.

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.

If you push the sled, as in part (a) of the figure, at an angle of 25°, what magnitude of force is needed to maintain the sled’s motion at a constant speed?

This problem explains how to calculate the force required to push a sled at a constant speed on snow, factoring in kinetic friction and an applied force at an angle of 25°. Learn how to break down the forces involved and use trigonometric functions to find the solution. Suitable for students in Grades 9-12.

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