Example: Two Objects

Question:

Two objects connected with a massless string slide on a frictionless surface and have masses m₁=5.00 kg and m₂=10.0 kg as shown below. The object on the right is pulled with a constant force of 10.0 N. What is the tension in the string?

Answer:

First a word about the terms used here: "massless" means simply that the object is so light that we can safely neglect its mass in our calculations. This is a fairly common occurrence in introductory physics. "Frictionless" means that we can neglect friction in our calculations.
Here we encounter a new force: string tension. It is shown in blue and denoted with the letter T in the figure. We use strings a lot because the direction of the force lies along the string. The tension T is equal in magnitude at both ends but opposite in direction. To figure out how big T is, we again use the component method, but this time separately for both masses:

Making a free-body diagram of the forces on m₁ (basically just the left half of the above drawing):
$\sum$Fy = N₁ - m₁ g = 0 => N₁ = m₁g
(this was expected from last example)
$\sum$F_x = F - T = m₁ a_x = m₁ a =>
T = F - m₁ a

We still have an unknown, the acceleration a.

Now a free-body diagram of the forces on m₂:
$\sum$F_y = N₂ - m₂ g = 0 => N₂ = m₂ g
(no new useful information here)
$\sum$F_x = T = m₂ a_x = m₂a => T = m ₂ a
Now we can plug this in above to get:
F = (m₁ + m₂) a => \[ \rm \mathbf{a = \frac{F}{m_{1}+ m_{2}}} \]
Almost done:

\[ \rm \mathbf{T = m_{2}a = \frac{m_{2}F}{m_{1} + m_{2}} \nonumber\\ = \frac{10 kg \cdot 10 N}{5 kg + 10 kg} \nonumber\\ = 6.67 N} \]