## Moving Boxes

In the diagram below, a moving truck is parked, ready to be unloaded. Two 4 kg boxes are arranged so that Box 2 rests on top of Box 1. The boxes are situated at the very back of the storage space, adjacent to the partially opened loading door, which only allows Box 1 to be pulled out. If the kinetic friction between the boxes is 0.3, and the kinetic friction between Box 1 and the floor is 0.1, what is Box 1's acceleration when the movers apply a 35 N force to pull it out?

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__**Hint**

**Hint**

The problem can be broken down at the two areas where friction is applicable: Box 2 on Box 1, and Box 1 on the bottom floor surface.

$$$F_{friction}=\mu_k(mass)(acceleration)$$$

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__**Hint 2**

**Hint 2**

Combining both friction sections, we can evaluate the forces on Box 1:

$$$\sum Forces=mass \times acceleration$$$

The problem can be broken down at the two areas where friction is applicable. Let's analyze the upper frictional force due to Box 2 on Box 1:

$$$F_{friction}=\mu_k(mass)(acceleration)$$$

$$$F_{upper}=(0.3)(4kg)(9.8m/s^2)=11.76\:N$$$

Next, let's evaluate the lower frictional force on Box 1 due to the bottom floor surface:

$$$F_{lower}=(0.1)(4kg+4kg)(9.8m/s^2)=7.84\:N$$$

Combining both friction sections, we can evaluate the forces on Box 1:

$$$\sum F=ma$$$

$$$F-F_{upper}-F_{lower}=ma$$$

$$$35N-11.76N-7.84N=(4kg)a$$$

Finally to solve for acceleration:

$$$a=\frac{15N}{4kg}=3.85\:m/s^2$$$

$$$3.85\:m/s^2$$$