Problem 50 - University Physics with Modern Physics

Question

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, starting from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?

Step-by-Step Solution

Answer

The tool weighs 4.515 N on Newtonia and 14.715 N on Earth.

Step 1: Determine Horizontal Acceleration

We know that a 12.0 N force moves the tool 16.0 m in 2.00 s. We can use the kinematic equation \( x = \frac{1}{2} a t^2 \) to find the acceleration \( a \). Given \( x = 16.0 \) m and \( t = 2.00 \) s, substituting in yields:\[16.0 = \frac{1}{2} a (2.00)^2\]Solving for \( a \), we find:\[16.0 = 2a \Rightarrow a = 8.0 \, \text{m/s}^2\]

Step 2: Calculate Mass of the Tool

The force equation \( F = ma \) allows us to calculate the mass \( m \) using the horizontal force and acceleration. Given \( F = 12.0 \) N and \( a = 8.0 \, \text{m/s}^2 \):\[m = \frac{F}{a} = \frac{12.0}{8.0} = 1.5 \, \text{kg}\]

Step 3: Calculate Gravitational Acceleration on Newtonia

When the tool is released from 10.0 m high, it takes 2.58 s to fall. Using the kinematic equation \( y = \frac{1}{2} g t^2 \) with \( y = 10.0 \) m and \( t = 2.58 \) s, we solve for \( g \):\[10.0 = \frac{1}{2} g (2.58)^2\]Solving for \( g \), we get:\[10.0 = 3.3241 g \Rightarrow g = \frac{10.0}{3.3241} \approx 3.01 \, \text{m/s}^2\]

Step 4: Calculate Weight on Newtonia

The weight of the tool on Newtonia can be found using the formula \( W = mg \), where \( m = 1.5 \, \text{kg} \) and \( g = 3.01 \, \text{m/s}^2 \):\[W = 1.5 \times 3.01 = 4.515 \, \text{N}\]

Step 5: Calculate Weight on Earth

Using Earth's gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \), calculate the weight on Earth with \( m = 1.5 \, \text{kg} \):\[W = 1.5 \times 9.81 = 14.715 \, \text{N}\]

Key Concepts

Gravitational Acceleration

Gravitational acceleration is a fundamental concept when studying the motion of objects under the influence of gravity. On Earth, this is denoted as \( g = 9.81 \, \text{m/s}^2 \). However, on other planets, the value of \( g \) can differ significantly. For example, in the problem of Newtonia, the gravitational acceleration is calculated to be approximately \( 3.01 \, \text{m/s}^2 \). This lower value indicates a weaker gravitational pull compared to Earth.
Understanding gravitational acceleration is crucial, as it determines how fast an object will accelerate towards the ground when it is dropped. The formula \( y = \frac{1}{2} g t^2 \) helps calculate the gravitational pull of different celestial bodies by measuring how long it takes for objects to fall a known distance.
When studying physics, comparing gravitational acceleration between planets can help explain phenomena such as why objects feel lighter or heavier. A lower \( g \) means objects fall slower and can appear lighter, a key insight when considering planetary exploration.

Kinematic Equations

Kinematic equations are equations of motion that help describe the relationship between an object's position, velocity, acceleration, and time. They form a vital part of solving physics problems related to motion. In the context of the exercise, the equation used is \( x = \frac{1}{2} a t^2 \), which helps calculate the acceleration of an object when its initial velocity is zero.
These equations are essential for solving problems where you have known quantities like force, distance, and time, and need to find unknowns such as acceleration or final velocity. The kinematic equation for distance, \( x = \frac{1}{2} a t^2 \), assumes constant acceleration and is particularly useful in experiments with frictionless surfaces.
Mastering kinematic equations allows students to predict how objects will move under various forces and accelerations. This knowledge is not only applicable in academic physics problems but also in real-world scenarios, such as car braking distances and flight trajectory planning.

Mass and Weight

Mass and weight, though often used interchangeably, are distinct concepts in physics. Mass is a measure of the amount of matter in an object, expressed in kilograms, and remains constant regardless of location. In our exercise, the tool's mass was calculated as \( 1.5 \, \text{kg} \).
Weight, however, is the force exerted by gravity on that mass and varies depending on the gravitational acceleration of the planet you are on. This relationship is expressed by the equation \( W = mg \), where \( W \) is weight in newtons, \( m \) is mass, and \( g \) is gravitational acceleration.
The problem illustrates this distinction well, showing that the tool's weight on Newtonia is approximately \( 4.515 \, \text{N} \) compared to \( 14.715 \, \text{N} \) on Earth. This highlights how an object's weight is dependent on local gravitational acceleration, reinforcing the need to consider both mass and gravitational forces for accurate calculations.

Physics Problem-Solving

Physics problem-solving often involves taking a systematic approach to identify and calculate various forces and motion properties, utilizing an array of mathematics and equations. The exercise on Newtonia provides a perfect example of this practice.
Breaking down the problem step-by-step is crucial:

• Begin by identifying known variables and conditions, such as force, distance, and time.
• Choose appropriate mathematical formulas, like kinematic equations, to find unknowns such as acceleration or gravitational force.
• Use logical reasoning and calculation, progressively solving for information such as mass and weight under differing gravitational conditions.

These steps ensure solutions are robust and comprehensible, aiding not only in understanding the problem at hand but also in applying the solutions to wider contexts. Practicing these problem-solving skills will aid students in tackling more complex physics scenarios with confidence and precision.