Questions: In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed 700 kN (about 71 tons). Assume that he pulled with a constant force that was 2.1 times his body weight, at an upward angle θ of 32 degrees from the horizontal. His mass was 81 kg, and he moved the cars by 1.0 m. Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.

Solution Steps

First, we need to calculate the force exerted by John Massis. Given that he pulled with a force that was 2.1 times his body weight, we can find this force as follows:

\[ F_{\text{pull}} = 2.1 \times (81 \, \text{kg} \times 9.81 \, \text{m/s}^2) \]

\[ F_{\text{pull}} = 2.1 \times 794.61 \, \text{N} \]

\[ F_{\text{pull}} = 1668.681 \, \text{N} \]

Next, we resolve this force into horizontal and vertical components. The horizontal component \( F_x \) is given by:

\[ F_x = F_{\text{pull}} \cos(32^\circ) \]

\[ F_x = 1668.681 \, \text{N} \times \cos(32^\circ) \]

\[ F_x \approx 1414.7 \, \text{N} \]

The work done \( W \) by the horizontal force over a distance of 1.0 m is:

\[ W = F_x \times d \]

\[ W = 1414.7 \, \text{N} \times 1.0 \, \text{m} \]

\[ W = 1414.7 \, \text{J} \]

The total mass of the cars is given by:

\[ m_{\text{cars}} = \frac{700 \, \text{kN}}{9.81 \, \text{m/s}^2} \]

\[ m_{\text{cars}} \approx 71355.8 \, \text{kg} \]

Using the work-energy principle, the work done is equal to the change in kinetic energy:

\[ W = \frac{1}{2} m_{\text{cars}} v^2 \]

Solving for \( v \):

\[ 1414.7 \, \text{J} = \frac{1}{2} \times 71355.8 \, \text{kg} \times v^2 \]

\[ v^2 = \frac{1414.7 \, \text{J}}{0.5 \times 71355.8 \, \text{kg}} \]

\[ v^2 \approx 0.0396 \, \text{m}^2/\text{s}^2 \]

\[ v \approx 0.1990 \, \text{m/s} \]

Final Answer