A 2.5 kg cart with a vertical front is pushed from behind across a horizontal surface by force F⃗. Rolling friction is negligible. A 100 g button is placed against the front of the cart. What minimum magnitude must the pushing force have to prevent the button from sliding down? The coefficient of static friction between the button and the cart is 0.55.
Express your answer with the appropriate units.

To solve this problem, we need to find the minimum force \( F \) that must be applied to prevent the button from sliding down the cart. Here's the step-by-step process:

### Known values:
- Mass of the cart: \( m_{\text{cart}} = 2.5 \, \text{kg} \)
- Mass of the button: \( m_{\text{button}} = 0.100 \, \text{kg} \)
- Coefficient of static friction between the button and the cart: \( \mu_s = 0.55 \)
- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)

### Key idea:
The static friction between the button and the cart is what prevents the button from sliding down. The frictional force \( f_s \) must balance the gravitational force acting on the button in the vertical direction.

The frictional force is given by:
\[
f_s = \mu_s \cdot N
\]
where \( N \) is the normal force exerted by the cart on the button.

The normal force in this case arises from the horizontal acceleration of the cart, which pushes the button against the front of the cart. The normal force is related to the horizontal acceleration \( a \) of the cart by Newton's second law:
\[
N = m_{\text{button}} \cdot a
\]

The button will not slide down if the static friction is at least as large as the gravitational force acting on the button:
\[
f_s \geq m_{\text{button}} \cdot g
\]

Thus, we set up the equation:
\[
\mu_s \cdot N = m_{\text{button}} \cdot g
\]

Substitute \( N = m_{\text{button}} \cdot a \) into this equation:
\[
\mu_s \cdot m_{\text{button}} \cdot a = m_{\text{button}} \cdot g
\]

The mass of the button cancels out:
\[
\mu_s \cdot a = g
\]

Solve for \( a \):
\[
a = \frac{g}{\mu_s}
\]

Substitute the known values:
\[
a = \frac{9.8 \, \text{m/s}^2}{0.55} = 17.82 \, \text{m/s}^2
\]

Now, we use Newton's second law to find the force \( F \) that must be applied to the cart to give it this acceleration:
\[
F = m_{\text{cart}} \cdot a
\]

Substitute the known values:
\[
F = 2.5 \, \text{kg} \cdot 17.82 \, \text{m/s}^2 = 44.55 \, \text{N}
\]

Thus, the minimum magnitude of the pushing force is:
\[
F = 44.6 \, \text{N}
\]

### Final Answer:
The minimum magnitude of the pushing force \( F \) is \( \boxed{44.6 \, \text{N}} \).

Would you like further details on any step or have questions?

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Here are 5 related questions:
1. How would the minimum force change if the button's mass was doubled?
2. What happens if the coefficient of static friction is reduced to 0.3?
3. How is the situation affected if the surface has a rough texture (introducing rolling friction)?
4. If the cart was moving up an incline, how would you modify the calculations?
5. What is the impact of increasing the cart's mass on the minimum force?

### Tip:
Always check the directions of the forces involved and the nature of the friction (static or kinetic) when solving physics problems involving motion and contact surfaces.

A 2.5 kg cart with a vertical front is pushed from behind across a horizontal surface by force F⃗. Rolling friction is negligible. A 100 g button is placed against the front of the cart. What minimum magnitude must the pushing force have to prevent the button from sliding down? The coefficient of static friction between the button and the cart is 0.55. Express your answer with the appropriate units.

Find the minimum pushing force needed to prevent a button from sliding down a 2.5 kg cart. This physics problem involves Newton's Second Law, static friction, and acceleration, using the coefficient of static friction (0.55) between the button and cart. Suitable for high school students, Grades 9-12.

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