Problem 35 - College Physics

Question

A block of ice with a mass of \(2.50 \mathrm{~kg}\) is moving on a frictionless, horizontal surface. At \(t=0,\) the block is moving to the right with a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\). Calculate the magnitude and direction of the velocity of the block after each of the following forces has been applied for \(5.00 \mathrm{~s}\) : (a) a force of \(5.00 \mathrm{~N}\) directed to the right; (b) a force of \(7.00 \mathrm{~N}\) directed to the left.

Step-by-Step Solution

Answer

(a) 18.00 m/s to the right; (b) 6.00 m/s to the left.

Step 1: Calculate Acceleration for Force (a)

First, find the acceleration imparted by the force of 5.00 N to the right. Use Newton's second law, given by the formula: \[ F = ma \]Solving for acceleration \( a \), we have:\[ a = \frac{F}{m} = \frac{5.00 \text{ N}}{2.50 \text{ kg}} = 2.00 \text{ m/s}^2 \].

Step 2: Calculate Final Velocity for Force (a)

Next, calculate the final velocity of the block after 5.00 s, using the equation for uniformly accelerated motion:\[ v = u + at \]Where:- \( v \) is the final velocity- \( u = 8.00 \text{ m/s} \) is the initial velocity- \( a = 2.00 \text{ m/s}^2 \) is the acceleration- \( t = 5.00 \text{ s} \) is the timePlugging in the values:\[ v = 8.00 \text{ m/s} + (2.00 \text{ m/s}^2 \times 5.00 \text{ s}) = 18.00 \text{ m/s} \].The velocity is directed to the right.

Step 3: Calculate Acceleration for Force (b)

For the force of 7.00 N directed to the left, first find the acceleration:\[ a = \frac{F}{m} = \frac{7.00 \text{ N}}{2.50 \text{ kg}} = 2.80 \text{ m/s}^2 \].Note that the acceleration is to the left because the force is to the left.

Step 4: Calculate Final Velocity for Force (b)

Find the final velocity after applying the leftward force for 5.00 s:\[ v = u + at \]Where the values are:- \( u = 8.00 \text{ m/s} \)- \( a = -2.80 \text{ m/s}^2 \) (negative because the acceleration is in the opposite direction)- \( t = 5.00 \text{ s} \)Enter the known values:\[ v = 8.00 \text{ m/s} + (-2.80 \text{ m/s}^2 \times 5.00 \text{ s}) = -6.00 \text{ m/s} \]. The negative sign indicates the direction is to the left.

Key Concepts

Kinematics

Kinematics is a branch of physics that studies motion without considering the forces that cause this motion. It focuses on describing motion in terms of objects moving from one point to another. The key elements of kinematics include displacement, velocity, and acceleration, providing crucial insights into how objects move.
In our exercise, the block of ice on a frictionless horizontal surface is a perfect illustration of a kinematics scenario. The initial conditions include:

• Initial velocity: the speed and direction of the block at the start, which is given as 8.00 m/s to the right.
• Time duration: the period during which forces are applied, noted as 5 seconds.

Understanding kinematics helps us predict future motion based on these initial conditions.

Velocity Calculation

Velocity is a vector quantity, meaning it includes both magnitude and direction. In the context of our exercise, calculating the final velocity of the block after a force is applied involves understanding how speed and direction change over time.
The formula used is:
\( v = u + at \),where:

• \( v \) is the final velocity that we are looking to calculate.
• \( u \) is the initial velocity, which is set at 8.00 m/s.
• \( a \) is the acceleration caused by the applied force.
• \( t \) is the time during which the force is applied, noted as 5.00 seconds.

This equation allows us to incorporate both the change in speed and the direction the block is moving in over time.

Acceleration

Acceleration is the rate of change of velocity with respect to time. It indicates how quickly an object is speeding up or slowing down. According to Newton's Second Law, acceleration is caused by a net force acting upon an object, and is calculated using the formula:
\[ a = \frac{F}{m} \],where:

• \( F \) is the force applied.
• \( m \) is the mass of the object.

The sign of acceleration depends on the direction of the force. In our exercise:

• For a 5.00 N force to the right: This results in a positive acceleration of 2.00 m/s².
• For a 7.00 N force to the left: This results in a negative acceleration of 2.80 m/s², reflecting a change in direction.

These calculations demonstrate how different forces lead to varying accelerations, impacting the object's velocity.