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We are given a system where a small box of mass \( m = 2.4 \, \text{kg} \) is placed on top of a larger box of mass \( 2m = 4.8 \, \text{kg} \). A force of \( F = 62.6 \, \text{N} \) is applied to the larger box, and both boxes accelerate together with the same acceleration. The coefficient of friction between all surfaces is \( \mu = 0.4 \). We are asked to determine the force accelerating the smaller box.

### Step 1: Total System Mass and Acceleration
First, let's find the total mass of the system:
\[
M_{\text{total}} = m + 2m = 2.4 \, \text{kg} + 4.8 \, \text{kg} = 7.2 \, \text{kg}
\]

Using Newton's second law, we can find the acceleration of the system:
\[
F = M_{\text{total}} \cdot a
\]
\[
a = \frac{F}{M_{\text{total}}} = \frac{62.6 \, \text{N}}{7.2 \, \text{kg}} = 8.694 \, \text{m/s}^2
\]

### Step 2: Friction Force Between the Boxes
The smaller box remains on top of the larger box without slipping due to friction. The frictional force provides the necessary horizontal force to accelerate the smaller box.

The normal force between the two boxes is equal to the weight of the smaller box:
\[
N = m \cdot g = 2.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 23.52 \, \text{N}
\]

The frictional force \( f_{\text{friction}} \) is given by:
\[
f_{\text{friction}} = \mu \cdot N = 0.4 \times 23.52 \, \text{N} = 9.408 \, \text{N}
\]

### Step 3: Force Accelerating the Smaller Box
The frictional force is responsible for accelerating the smaller box, and this force is what prevents the smaller box from slipping off the larger one. Since the friction force is the only horizontal force acting on the smaller box, it must equal the force accelerating the smaller box.

Thus, the force accelerating the smaller box is:
\[
F_{\text{small}} = f_{\text{friction}} = 9.408 \, \text{N}
\]

### Final Answer:
The force accelerating the smaller box is approximately **9.41 N**.

Would you like further details or clarifications? Here are some related questions for deeper understanding:

1. How would the acceleration change if the applied force \( F \) increased?
2. What would happen if the coefficient of friction were lower than 0.4?
3. How would the system behave if the surfaces were frictionless?
4. Could the friction force ever exceed its current value of 9.408 N?
5. How would the problem change if the masses of the boxes were reversed?

**Tip:** Whenever dealing with systems involving friction, remember that frictional force is limited by the normal force and the coefficient of friction.

A small box of mass 2.4kg is placed on top of a larger box with double the mass as shown in the diagram. When a force of 62.6N is applied to the large box, both boxes accelerate to the right with the same acceleration. If the coefficient of friction between all surfaces is 0.4, what would be the force accelerating the smaller mass?

In this physics problem, explore how a small box of mass 2.4kg and a larger box of mass 4.8kg accelerate together when a force of 62.6N is applied to the larger box. The solution involves Newton's second law, friction force calculations, and acceleration. Suitable for high school physics students.

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