Problem 11 - University Physics with Modern Physics

Question

A hockey puck with mass 0.160 \(\mathrm{kg}\) 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 \(\mathrm{N}\) to the puck, parallel to the \(x\) -axis; he continues to apply this force until \(t=2.00 \mathrm{s}\) . (a) What are the position and speed of the puck at \(t=2.00 \mathrm{s?}\) (b) If the same force is again applied at \(t=5.00 \mathrm{s}\) , what are position and speed of the puck at \(t=7.00 \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) Velocity = 3.125 m/s, Position = 3.125 m; (b) Velocity = 6.250 m/s, Position = 21.875 m.

Step 1: Determine the acceleration

To find the acceleration of the puck, we use Newton's second law, \( F = ma \), where \( F \) is the force applied, \( m \) is the mass of the puck, and \( a \) is the acceleration. Given \( F = 0.250 \, \mathrm{N} \) and \( m = 0.160 \, \mathrm{kg} \), we calculate:\[ a = \frac{F}{m} = \frac{0.250}{0.160} = 1.5625 \, \mathrm{m/s}^2 \]

Step 2: Find velocity at t = 2.00 s

Since the puck starts from rest, its initial velocity \( u = 0 \). The final velocity \( v \) can be calculated using the kinematic equation: \( v = u + at \). Substitute \( a = 1.5625 \, \mathrm{m/s}^2 \) and \( t = 2.00 \, \mathrm{s} \):\[ v = 0 + (1.5625)(2.00) = 3.125 \, \mathrm{m/s} \]

Step 3: Calculate position at t = 2.00 s

Use the kinematic equation for position: \( s = ut + \frac{1}{2}at^2 \). With \( u = 0 \), \( a = 1.5625 \, \mathrm{m/s}^2 \), and \( t = 2.00 \, \mathrm{s} \):\[ s = 0 + \frac{1}{2}(1.5625)(2.00)^2 = 3.125 \, \mathrm{m} \]

Step 4: Establish Initial Conditions for the Second Interval

At \( t = 5.00 \, \mathrm{s} \), the puck continues to move without additional force from \( t = 2.00 \) to \( t = 5.00 \). During this interval, it maintains a constant velocity of \( 3.125 \, \mathrm{m/s} \). Calculate the distance traveled during these 3 seconds (\( 5.00 \, \mathrm{s} - 2.00 \, \mathrm{s} \)) using \( s = vt \):\[ s = 3.125 \times 3.00 = 9.375 \, \mathrm{m} \]Including the previous displacement, total position at \( t = 5.00 \) is:\[ 9.375 \, \mathrm{m} + 3.125 \, \mathrm{m} = 12.500 \, \mathrm{m} \]

Step 5: Apply Force Again at t = 5.00 s

From \( t = 5.00 \) to \( t = 7.00 \) seconds, the force is reapplied, producing the same acceleration of \( 1.5625 \, \mathrm{m/s}^2 \). We recompute velocity at \( t = 7.00 \) using \( v = u + at \) with \( u = 3.125 \, \mathrm{m/s} \):\[ v = 3.125 + (1.5625)(2.00) = 6.250 \, \mathrm{m/s} \]

Step 6: Find Position at t = 7.00 s

Calculate additional displacement from \( t = 5.00 \) to \( t = 7.00 \) using \( s = ut + \frac{1}{2}at^2 \) with \( u = 3.125 \, \mathrm{m/s} \), \( t = 2.00 \) and \( a = 1.5625 \, \mathrm{m/s}^2 \):\[ s = (3.125)(2.00) + \frac{1}{2}(1.5625)(2.00)^2 = 6.250 + 3.125 = 9.375 \, \mathrm{m} \]

Step 7: Calculate Total Position at t = 7.00 s

Add the displacement from previous steps to find the total position at \( t = 7.00 \):\[ 12.500 \, \mathrm{m} + 9.375 \, \mathrm{m} = 21.875 \, \mathrm{m} \]

Key Concepts

Kinematic Equations

Kinematic equations are essential when analyzing motion. They help us calculate positions, velocities, and accelerations of objects in motion over time when acceleration is constant. In the exercise, we use two key kinematic equations.

• The first is \( v = u + at \), which finds the final velocity \( v \) of an object that started with initial velocity \( u \) and accelerates over time \( t \) at rate \( a \).
• The second is the position equation \( s = ut + \frac{1}{2}at^2 \). This calculates the displacement \( s \) of an object.

Both equations are particularly useful when solving problems involving constant acceleration, as in this case where a hockey puck is pushed across a frictionless surface. Understanding these equations is critical for predicting the motion's outcome based on initial conditions and applied forces.
They offer a direct way to connect the motion's factors, paving the path for more accurate predictions.

Constant Acceleration

Constant acceleration is a scenario where the rate of change of velocity remains the same over a time period. This is especially common in simplified physics models, such as the hockey puck on a frictionless surface. Given that the force applied to the puck is constant, the acceleration is also constant, which simplifies calculations.When you know the acceleration is constant:

• Predicting future motion becomes easier.
• You can rely on linear kinematic equations like \( v = u + at \) and \( s = ut + \frac{1}{2}at^2 \).

In this exercise, the constant acceleration ensures that calculations based on these equations can accurately describe the motion. Knowing the constant force, the mass of the puck, and using Newton's second law, you can determine this unchanging acceleration.

Force and Motion

Force and motion are deeply intertwined concepts in physics, governed by Newton's Laws. The second law, \( F = ma \), is particularly relevant here. It tells us how force affects motion:

• Force \( F \) applied to an object with mass \( m \) results in acceleration \( a \).
• This relationship is used to predict how an object starts moving or changes its velocity.

In the exercise, a 0.250 N force is applied to a puck with a mass of 0.160 kg. According to the formula, this gives us an acceleration of 1.5625 m/s². Understanding this principle helps explain how motion begins and evolves when a force is applied, helping students see the direct relationship between force, mass, and acceleration.

Frictionless Surfaces

A frictionless surface offers a simplified model to study motion without the interference of forces that oppose movement, such as friction.

• This idealized condition allows us to focus on the effects of other forces, like applied force, in isolation.
• It makes calculations more straightforward, as you need not account for velocity loss due to friction.

In this exercise, a hockey rink is assumed to be frictionless, ensuring pure application of Newton's laws without additional forces. This offers a clearer understanding of force and motion dynamics. By neglecting friction, students can grasp core principles more easily, paving the way for handling more complex problems where friction is a factor. The simplicity of frictionless systems is a powerful educational tool in learning physics basics.