Problem 11 - College Physics

Question

A dock worker applies a constant horizontal force of 80.0 \(\mathrm{N}\) to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 \(\mathrm{m}\) in the first 5.00 \(\mathrm{s}\) s. What is the mass of the block of ice?

Step-by-Step Solution

Answer

The mass of the block of ice is approximately 90.91 kg.

Step 1: Identify the Known Values

The force applied, \( F = 80.0 \, \mathrm{N} \). The block moves a distance of \( s = 11.0 \, \mathrm{m} \) in time \( t = 5.00 \, \mathrm{s} \). Since the block starts from rest, the initial velocity \( u = 0 \, \mathrm{m/s} \).

Step 2: Use Kinematic Equation to Find Acceleration

The kinematic equation \( s = ut + \frac{1}{2}a t^2 \) is used to find the acceleration \( a \). Substituting the values: \( 11.0 = 0 + \frac{1}{2}a (5.00)^2 \). Simplifying, we have \( 11.0 = \frac{1}{2} a \times 25 \), so \( a = \frac{22}{25} = 0.88 \mathrm{m/s^2} \).

Step 3: Use Newton's Second Law to Find Mass

According to Newton's second law, \( F = ma \). Rearrange this equation to solve for mass \( m \): \( m = \frac{F}{a} \). Substitute the known values: \( m = \frac{80.0 \, \mathrm{N}}{0.88 \, \mathrm{m/s^2}} \). Calculating gives \( m \approx 90.91 \, \mathrm{kg} \).

Key Concepts

Kinematic Equations

Kinematic equations are a crucial part of physics, especially in problems involving motion. They help in describing the motion of objects, providing a relationship between variables like displacement, velocity, acceleration, and time. In this exercise, the kinematic equation used is:

• \( s = ut + \frac{1}{2} a t^2 \)

This equation relates the displacement \( s \) of an object, its initial velocity \( u \), the acceleration \( a \), and the time \( t \). Since the block of ice starts from rest, the initial velocity \( u = 0 \). By plugging in the known values, we calculate the acceleration \( a \) of the block. The equation simplifies to \( 11.0 = \frac{1}{2} a (5.00)^2 \), which gives us \( a = 0.88 \, \mathrm{m/s^2} \). Understanding how to manipulate these equations allows for solving complex motion problems.

Constant Force

Applying a constant force to an object leads to a uniform change in motion, provided there is no opposing force like friction. In this exercise, a constant force of 80.0 \( \mathrm{N} \) is applied to the block of ice. Newton's Laws tell us how this constant force affects the motion.

• Since this force is constant and horizontal, and friction is negligible, all of the force contributes to moving the block.
• The constant force ensures a constant acceleration, assuming mass stays the same.

Thus, the block of ice experiences a steady increase in speed, driven purely by the applied force.

Acceleration Calculation

Acceleration is the rate of change of velocity and is a critical factor in determining how quickly an object speeds up or slows down. To find acceleration in this scenario, we used the kinematic equation:

• \( s = ut + \frac{1}{2} a t^2 \)

We substituted the values to get \( a = 0.88 \, \mathrm{m/s^2} \). This means that every second, the velocity of the block increases by 0.88 meters per second. Acceleration here is constant because the force applied is consistent and there is no friction to alter the motion. Calculating acceleration accurately is essential as it affects the final calculation of mass through Newton's Second Law.

Mass Determination

Mass determination involves understanding the relationship between force, mass, and acceleration, as expressed in Newton's Second Law of Motion:

• \( F = ma \)

Here, \( F \) is the force applied, \( m \) is the mass, and \( a \) is acceleration. We rearrange the equation to find the mass:

• \( m = \frac{F}{a} \)

By substituting the known values into the equation, \( m = \frac{80.0 \, \mathrm{N}}{0.88 \, \mathrm{m/s^2}} \), we calculate that the mass of the block is approximately \( 90.91 \, \mathrm{kg} \). Mass is a fundamental property, and accurately determining it allows us to understand how much matter is present in the block of ice, influencing how it responds to the applied force.