Problem 47 - University Physics with Modern Physics

Question

A 6.50-kg instrument is hanging by a vertical wire inside a space ship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 276 \(\mathrm{m}\) in 15.0 s with constant acceleration. (a) Draw a free-body diagram for the instrument during this time. Indicate which force is greater. (b) Find the force that the wire exerts on the instrument.

Step-by-Step Solution

Answer

The wire exerts a force of approximately 79.63 N on the instrument.

Step 1: Analyze Forces on the Instrument

The instrument is subject to two forces: 1. The gravitational force - Its magnitude is calculated using the equation: \[ F_g = m \times g \] - where \( m = 6.50 \, \text{kg} \) is the mass of the instrument, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. 2. The tension force from the wire, \( T \), which acts upward in response to the combined effects of gravity and the spaceship's acceleration.

Step 2: Draw the Free-Body Diagram

Draw a vertical arrow pointing downward labeled \( F_g \) (the force due to gravity) and a vertical arrow pointing upward labeled \( T \) (the tension force from the wire). The length of the arrow representing tension will be longer than the arrow for gravity because the spaceship is accelerating upwards, indicating that \( T > F_g \).

Step 3: Calculate the Net Acceleration of the Spaceship

The spaceship starts from rest and reaches a final altitude of 276 m in 15.0 s. Use the kinematic equation:\[ s = ut + \frac{1}{2}at^2 \]Where: \( s = 276 \, \text{m} \), \( u = 0 \), \( t = 15.0 \, \text{s} \), and \( a \) is the unknown acceleration. Substitute the known values to find \( a \):\[ 276 = 0 \times 15 + \frac{1}{2}a \times (15.0)^2 \]\[ 276 = \frac{1}{2}a \times 225 \]\[ 552 = 225a \]\[ a = \frac{552}{225} = 2.45 \, \text{m/s}^2 \].

Step 4: Calculate the Total Force Exerted by the Wire

The net force on the instrument due to the combined effects of the spaceship's acceleration and gravity is calculated using Newton's second law of motion:\[ F_{ ext{net}} = m \times a_{ ext{net}} = T - F_g \]Rearranging gives \( T = F_{ ext{net}} + F_g \). Calculate \( F_g \):\[ F_g = 6.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 63.7 \, \text{N} \].Calculate the net force from acceleration, \( F_{ ext{net}} \):\[ F_{ ext{net}} = m \times a = 6.50 \, \text{kg} \times 2.45 \, \text{m/s}^2 = 15.925 \, \text{N} \].Now find \( T \):\[ T = 15.925 \, \text{N} + 63.7 \, \text{N} = 79.625 \, \text{N} \].

Key Concepts

Newton's second law

Newton's second law of motion is a cornerstone of classical mechanics. It tells us how the motion of an object changes when it is subjected to forces. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, which can be represented mathematically as \( F = m \cdot a \). This simple equation explains how the speed and direction of an object change.In the given problem, the instrument inside the spaceship experiences forces that cause it to accelerate. We use this law to understand the relationship between the forces acting on the instrument and its motion. The tension in the wire has to counteract both the gravitational pull (weight) and impart an additional force to provide the necessary acceleration. Let's break it down:

• Gravity exerts a downward force, calculated as the product of the mass and the gravitational acceleration (\( 9.8 \, \text{m/s}^2 \)).
• The wire must provide an upward tension force which is larger than the gravitational force to provide upward acceleration.

This results in the equation \( T - F_g = m \times a_{\text{net}} \), allowing us to solve for \( T \), the tension in the wire.

Forces and motion

Forces and motion are directly related concepts in physics, meaning that a force will change an object's state of motion. In this exercise, understanding these two interrelated ideas is crucial when analyzing the spaceship's journey and the instrument's resulting force dynamics.As the spaceship travels upward:

• The wire has to exert a force greater than its weight to provide upward acceleration.
• The gravitational force continues to act downward, unaffected by the spaceship's motion.
• The spaceship's engines create an upward force, which indirectly affects the instrument through the tension in the wire.

The spaceship's ascent manifests constant acceleration, which is seen through the distance covered—276 meters in 15 seconds. Using kinematics, we calculate this as \( 2.45 \, \text{m/s}^2 \). This information is vital for solving the problem because it helps determine the net force, giving insights into how different forces interact to change motion.The takeaway is that forces change motion, and understanding this concept allows us to predict and calculate the outcomes like tension in the wire during the spaceship's lift-off.

Free-body diagrams

A free-body diagram is an essential tool in physics to visualize forces acting on an object. It simplifies analyzing forces by representing each force as an arrow pointing in the direction the force is exerted. The length of each arrow is proportional to the force's magnitude.In the scenario provided:

• The gravitational force \( F_g \) is shown with an arrow pointing downward.
• The tension \( T \), which is the force exerted by the wire, is represented with an upward arrow, longer than the gravitational force arrow, indicating its greater magnitude.

This visual tool clarifies how forces combine to affect motion, making complex problems easier to understand and solve.By drawing out these forces, we can quickly see:

• Tension in the wire exceeds the gravitational force, providing the necessary upward acceleration.
• The net force acting on the instrument aligns with the direction of acceleration.

In conclusion, free-body diagrams are indispensable for dissecting forces and motion, allowing a clear visualization which aids in comprehending the dynamics at play.