Questions: A loaded tractor-trailer with a total mass of 5400 kg traveling at 2.7 km/h hits a loading dock and comes to a stop in 0.58 s. What is the magnitude of the average force exerted on the truck by the dock? Express your answer using two significant figures. F= N

Solution Steps

First, we need to convert the initial velocity from km/h to m/s.

\[ 2.7 \, \text{km/h} = 2.7 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 0.75 \, \text{m/s} \]

Using the formula for acceleration \(a\):

\[ a = \frac{\Delta v}{\Delta t} = \frac{0 - 0.75 \, \text{m/s}}{0.58 \, \text{s}} = -1.2931 \, \text{m/s}^2 \]

Using Newton's second law \(F = ma\):

\[ F = 5400 \, \text{kg} \times (-1.2931 \, \text{m/s}^2) = -6982.74 \, \text{N} \]

Since we are asked for the magnitude of the force:

\[ |F| = 6982.74 \, \text{N} \]

Rounding the force to two significant figures:

\[ |F| \approx 7000 \, \text{N} \]

Final Answer