How is the distance 23m?

Question

1. Two boxes of mass 30.0 kg and 10.0 kg are at rest side by side (Figure 5). You apply a force of 3.0* 10*2 N on the first box for 5.0 s, and they both slide across the floor. The larger box has a force friction of 180N, and the smaller box 60N. Question: Calculate the total distance travelled by the boxes.

Answer 1

I have to assume that the 30 kg box is being pushed on since there is no diagram. It actually doesn't matter because the force of friction is proportional to their masses (also weights), so the coefficient of friction is the same and they will stay together as a unit.

Therefore, you can act as if a mass of 40 kg experiences an applied force of 300 N (assuming 3.0 x 10²) and a friction force of 240N. The net force is 300-240 = 60 N

The acceleration a = F_net/m_T = 60N/40 kg = 1.5 m/s² for 5 seconds, then the boxes experience a force of 240N/40 kg = -6 m/s² until they stop (There are many ways to solve this, but a fast way is to realize that the change in velocity is the same magnitude so that Δt = Δv/a , they will take 5/4 = 1.25 seconds to slow back down to v = 0)

d = 1/2at² for each section (In the second part, the same distance is traveled going from 0 to a speed as going from a speed to 0 with deceleration of the same magnitude)

d = 1/2 (1.5 m/s²)(5 s)² + 1/2 (6.0 m/s²) (1.25 s)² = 23.4 m

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