Answer
The free-body diagram of the cars is shown on the right. The force exerted by John Massis is
$$
F=2.5 m g=2.5(80 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)=1960 \mathrm{N}
$$
since the motion is along the horizontal $x$ -axis, using Newton's second law, we have $F x=F \cos \theta=M a_{x},$ where $M$ is the total mass of the railroad cars. Thus, the acceleration of the cars is
$$
a_{x}=\frac{F \cos \theta}{M}=\frac{(1960 \mathrm{N}) \cos 30^{\circ}}{\left(7.0 \times 10^{5} \mathrm{N} / 9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=0.024 \mathrm{\ m} / \mathrm{s}^{2}
$$
Using Eq. $2-16,$ the speed of the car at the end of the pull is
$$
v_{x}=\sqrt{2 a_{x} \Delta x}=\sqrt{2\left(0.024 \mathrm{m} / \mathrm{s}^{2}\right)(1.0 \mathrm{m})}=0.22 \mathrm{\ m/s}
$$
Work Step by Step
The free-body diagram of the cars is shown on the right. The force exerted by John Massis is
$$
F=2.5 m g=2.5(80 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)=1960 \mathrm{N}
$$
since the motion is along the horizontal $x$ -axis, using Newton's second law, we have $F x=F \cos \theta=M a_{x},$ where $M$ is the total mass of the railroad cars. Thus, the acceleration of the cars is
$$
a_{x}=\frac{F \cos \theta}{M}=\frac{(1960 \mathrm{N}) \cos 30^{\circ}}{\left(7.0 \times 10^{5} \mathrm{N} / 9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=0.024 \mathrm{\ m} / \mathrm{s}^{2}
$$
Using Eq. $2-16,$ the speed of the car at the end of the pull is
$$
v_{x}=\sqrt{2 a_{x} \Delta x}=\sqrt{2\left(0.024 \mathrm{m} / \mathrm{s}^{2}\right)(1.0 \mathrm{m})}=0.22 \mathrm{\ m/s}
$$
