Question:

A 2 kg block is placed on a frictionless surface and is connected to a 5 kg hanging mass by a light string passing over a frictionless pulley, as shown in the diagram below. A force of 10 N is applied to the 2 kg block horizontally to the right. Find the acceleration of each mass and the tension in the string.

[Diagram: (The diagram shows a 2 kg block connected by a string passing over a pulley to a 5 kg hanging mass.)]

Answer:

First, we start by drawing the free-body diagrams for each mass.

For the 2 kg block:

• Force applied, F = 10 N to the right
• Tension in the string, T to the left
• Weight, W = 2 kg * 9.8 m/s^2 = 19.6 N downward
• Normal force, N = 19.6 N upward

For the 5 kg hanging mass:

• Tension in the string, T upward
• Weight, W = 5 kg * 9.8 m/s^2 = 49 N downward

Using Newton's second law, we can set up the following equations for the two masses:

For the 2 kg block: [ΣF_x = ma_x ] [10 N - T = 2a]

For the 5 kg hanging mass: [ΣF_y = ma_y ] [T = 5a]

Next, we can solve the equations using the two unknowns, a (acceleration) and T (tension).

First, we can solve for T by adding the equations for the 2 kg block and the 5 kg hanging mass: [10 - T = 2a] [T = 5a] Substitute the second equation into the first equation: [10 - 5a = 2a] [10 = 7a]

This gives us: [a = \frac{10}{7} , m/s^2]

After finding the acceleration, we can then find the tension in the string: [T = 5 \cdot \frac{10}{7} ] [T \approx7.14, N]

So, the acceleration of the 2 kg block is approximately $\frac{10}{7} \, m/s^2$ and the tension in the string is approximately 7.14 N.