Problem 19 - Physics

Question

Two blocks are connected by a wire that has a mass perunit length of \(8.50 \times 10^{-4} \mathrm{~kg} / \mathrm{m} .\) One block has a mass of \(19.0 \mathrm{~kg},\) and the other has a mass of 42.0 kg. These blocks are being pulled across a horizontal frictionless floor by a horizontal force \(\overrightarrow{\mathrm{P}}\) that is applied to the less massive block. A transverse wave travels on the wire between the blocks with a speed of \(352 \mathrm{~m} / \mathrm{s}\) (relative to the wire). The mass of the wire is negligible compared to the mass of the blocks. Find the magnitude of \(\overrightarrow{\mathbf{P}}\).

Step-by-Step Solution

Answer

The force required is 153.1 N.

Step 1: Calculate the total mass of the system

First, calculate the total mass of the entire system, which includes both blocks. The total mass is the sum of the masses of the two blocks.\[ m_{total} = m_1 + m_2 = 19.0 \, \text{kg} + 42.0 \, \text{kg} = 61.0 \, \text{kg} \]

Step 2: Calculate the tension in the wire

To find the tension in the wire, we use the formula relating wave speed on a string to tension and mass per unit length. The wave speed \(v\) on a string is given by:\[ v = \sqrt{\frac{T}{\mu}} \]where \(T\) is the tension in the wire and \(\mu\) is the mass per unit length.Rearrange this equation to solve for \(T\):\[ T = v^2 \times \mu \]Substitute the given values:\[ T = (352 \, \text{m/s})^2 \times 8.50 \times 10^{-4} \, \text{kg/m} = 105.4 \, \text{N} \]

Step 3: Determine the acceleration of the system

Since the tension in the wire is balanced by the force required to accelerate the 42.0 kg block, we use Newton's Second Law for this block to find the acceleration \(a\):\[ T = m_2 \cdot a \]Solve for \(a\):\[ a = \frac{T}{m_2} = \frac{105.4 \, N}{42.0 \, \text{kg}} = 2.51 \, \text{m/s}^2 \]

Step 4: Find the required force to move the entire system

Now apply Newton's Second Law to the entire system of two blocks to find the external force \(\overrightarrow{P}\), knowing the total system mass and acceleration:\[ \overrightarrow{P} = m_{total} \cdot a \]Substitute the values of \(m_{total}\) and \(a\):\[ \overrightarrow{P} = 61.0 \, \text{kg} \times 2.51 \, \text{m/s}^2 = 153.1 \, \text{N} \]

Step 5: Final Answer

Therefore, the magnitude of the horizontal force \(\overrightarrow{P}\) needed is \(153.1 \, \text{N}\).

Key Concepts

Tension in a String

When you hear about tension in a string, think of it as the pulling force exerted along the string. Imagine you have a stretchy band and you pull it between two hands; the force you feel pulling back at your hands is similar to the tension. In our problem, the tension arises because of the force required to accelerate the mass which the string connects.

Important things to note about tension in a string:

• It's uniform along the string if there's perfect symmetry and no external forces act aside from the pulling at the ends.
• It results from the interaction between the string's mass and acceleration, alongside any forces applied to it.

To calculate tension using a wave on the string, we use the formula:\[T = v^2 \times \mu\]where \(v\) is the wave speed, and \(\mu\) is the mass per unit length. This relationship is key to unraveling problems with waves on strings, linking physical properties of the string to measurable wave speeds.

Mass per Unit Length

Mass per unit length, typically symbolized by \(\mu\), is simply the mass of the string divided by its length. Think of it like the string's "density," describing how much mass is packed per meter.

A few points about mass per unit length:

• This factor contributes to how easy or hard it is to accelerate the string.
• A higher mass per unit length means the string can 'carry' more wave characteristics, like higher tensions for given wave speeds.

In our exercise, the given \(\mu\) is \(8.50 \times 10^{-4}\,\text{kg/m}\), which informs us about the string's capability to relate to the wave speed and thus determine necessary tension — key when connecting motion to force.

Transverse Wave Speed

Transverse wave speed is how fast a wave travels down a string, traditionally measured in meters per second (m/s). We visualize this by picturing a wave ripple moving along a string when it's shaken side to side. This speed is crucial for determining how quickly energy is transferred through the string.

Understanding transverse wave speed:

• The wave speed depends directly on the tension in the string and its mass per unit length.
• Faster waves indicate higher tension or lower mass per unit length.

The equation representing this relationship is:\[v = \sqrt{\frac{T}{\mu}}\]In this formula, \(T\) stands for tension, and \(\mu\) is our mass per unit length. In our specific problem, the wave speed is 352 m/s, and plugging this into the equation allows us to find the required tension in the string, linking this dynamic property directly to measurable conditions.