Question

A hockey puck with mass 0.160 kg is at rest at the origin (\(x =\) 0) on the horizontal, frictionless surface of the rink. At time \(t =\) 0 a player applies a force of 0.250 N to the puck, parallel to the \(x\)-axis; she continues to apply this force until \(t =\) 2.00 s. (a) What are the position and speed of the puck at \(t =\) 2.00 s? (b) If the same force is again applied at \(t =\) 5.00 s, what are the position and speed of the puck at \(t =\) 7.00 s?

Short answer

At 2.00 s, position is 3.125 m, speed is 3.125 m/s. At 7.00 s, position is 21.875 m, speed is 6.25 m/s.

Step-by-step solution

Understand the Problem

We have a hockey puck initially at rest. It's acted upon by a constant force along the x-axis, creating a frictionless scenario. We need to find its position and speed after 2 seconds of applying this force, and then check its state again at 7 seconds, after reapplying the same force at 5 seconds.

Calculate Acceleration

Using Newton's second law, the acceleration of the puck can be found using the equation: \[ a = \frac{F}{m} \]where \( F = 0.250 \) N and \( m = 0.160 \) kg. Plug in these values:\[ a = \frac{0.250}{0.160} = 1.5625 \text{ m/s}^2 \]

Calculate Velocity at t = 2.00 s

The velocity of the puck can be determined using the equation for constant acceleration:\[ v = u + at \]where \( u = 0 \) (initial velocity), \( a = 1.5625 \text{ m/s}^2 \), and \( t = 2.00 \) s. Therefore:\[ v = 0 + 1.5625 \times 2.00 = 3.125 \text{ m/s} \]

Calculate Position at t = 2.00 s

The position of the puck is given by the equation:\[ x = ut + \frac{1}{2}at^2 \]Substituting the known values \( u = 0 \), \( a = 1.5625 \text{ m/s}^2 \), and \( t = 2.00 \) s, we get:\[ x = 0 \times 2.00 + \frac{1}{2} \times 1.5625 \times (2.00)^2 = 3.125 \text{ m} \]

Resume Force Application at t = 5.00 s

The puck continues with constant velocity of 3.125 m/s from \( t = 2.00 \) s until \( t = 5.00 \) s. From \( t = 5.00 \) s, the same force is applied again.

Calculate Position and Velocity at t = 5.00 s

The puck moves at 3.125 m/s for 3 more seconds (from \( t = 2.00 \) to \( t = 5.00 \)). The additional distance travelled is:\[ x_{additional} = 3.125 \text{ m/s} \times 3 \text{ s} = 9.375 \text{ m} \]Thus, at \( t = 5.00 \) s, the current position \( x = 3.125 + 9.375 = 12.5 \text{ m} \) and velocity \( v = 3.125 \text{ m/s} \).

Calculate New Velocity at t = 7.00 s

From \( t = 5.00 \) s to \( t = 7.00 \) s, the force is re-applied, providing additional acceleration. Using \[ v = u + at \]with \( u = 3.125 \text{ m/s} \), \( a = 1.5625 \text{ m/s}^2 \), and \( t = 2.00 \) s:\[ v = 3.125 + 1.5625 \times 2.00 = 6.25 \text{ m/s} \]

Calculate New Position at t = 7.00 s

The position of the puck from \( t = 5.00 \) s to \( t = 7.00 \) s is:\[ x = u \times t + \frac{1}{2}at^2 \]With \( u = 3.125 \text{ m/s} \), \( a = 1.5625 \text{ m/s}^2 \), and \( t = 2.00 \) s:\[ x_{additional} = 3.125 \times 2.00 + \frac{1}{2} \times 1.5625 \times (2.00)^2 = 9.375 \text{ m} \]Adding the previous position (12.5 m), we get:\[ x = 12.5 + 6.25 = 21.875 \text{ m} \]

Key concepts

constant acceleration equations

When dealing with motion, especially with a scenario like a puck sliding on a frictionless surface, constant acceleration equations are invaluable. These equations help us predict how an object's velocity and position change over time when it experiences constant acceleration.
To find acceleration, we use Newton's Second Law of Motion, which states that the acceleration is the force applied divided by the mass of the object: \( a = \frac{F}{m} \).
Once we know the acceleration, we can calculate the velocity at any given time with the equation \( v = u + at \), where \( u \) is the initial velocity.

• This equation tells us that velocity depends linearly on time and directly on acceleration.
• The initial velocity \( u \) is assumed to be zero if the object starts from rest.

In addition, the equation \( x = ut + \frac{1}{2}at^2 \) determines the position over time, showing the relationship between distance, time, initial velocity, and acceleration.
These equations allow us to solve many problems related to moving objects efficiently under the influence of consistent forces.

kinematics

Kinematics is all about the description of motion. When a hockey puck slides on a surface, understanding its movement involves determining its position and velocity at different time intervals. Kinematics provides the framework to do this by breaking down motion into several components.
We start by identifying initial conditions such as the initial position and velocity. In our exercise, the puck starts at rest at the origin. This simplifies our calculations as the initial conditions are zero for this specific context.
Once a force is applied, the puck starts accelerating, and kinematics allows us to calculate how far it travels and at what speed it moves using the constant acceleration equations mentioned above. These simplified equations assume no outside resistances, like friction, and allow us to focus solely on the movement caused by the provided force.

• Kinematics assumes motion in a defined space usually represented on a set of axes, making it easier to apply mathematical formulas.
• It also provides visual insight into the object's path through position-time and velocity-time graphs, useful for data interpretation and prediction.

By examining the change in positions and velocities over various time intervals, students can grasp how objects behave under different force applications and movements.

force and motion

Force and motion are tightly interlinked concepts in physics, crucial for understanding how objects interact with one another.
Newton's Second Law, \( F = ma \), is essential here as it forms the backbone for predicting how an object reacts to a given force. In the exercise, the applied force initiates the puck's motion on the frictionless rink, providing a straightforward example of this relationship.

• This law tells us that the acceleration of an object is directly proportional to the force applied, given a constant mass, meaning more force means more acceleration.
• Motion depends significantly on the nature and direction of the applied force.

The concept of force and motion explains why the model of a frictionless surface is used. Removing friction means no resistance to the applied force, simplifying calculations and highlighting the pure relationship between force, mass, and motion.
Understanding force and motion is fundamental to mastering how various systems work, from sports dynamics to complex real-world machinery operations. Grasping these principles allows students to predict and analyze motion resulting from various forces effectively.