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Chapter 4: Problem 11

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 the position and speed of the puck at \(t=7.00 \mathrm{s} ?\)

Short Answer

Expert verified
At t=2.00 s, x=3.125 m, v=3.125 m/s. At t=7.00 s, x=20.625 m, v=6.25 m/s.

Step by step solution

01

Calculate the acceleration

Using Newton's second law, calculate the acceleration of the puck. The formula is: \[ F = m \cdot a \]Where:- \( F = 0.250\, \mathrm{N} \) (force applied)- \( m = 0.160\, \mathrm{kg} \) (mass of the puck)Rearrange the equation to solve for acceleration \( a \):\[ a = \frac{F}{m} = \frac{0.250}{0.160} = 1.5625\, \mathrm{m/s^2} \]
02

Find the velocity at t=2.00 s

Since the puck starts from rest, its initial velocity \( v_0 = 0 \). Use the formula for velocity under constant acceleration:\[ v = v_0 + a \cdot t \]Substitute the values:\[ v = 0 + 1.5625 \times 2.00 = 3.125\, \mathrm{m/s} \]
03

Calculate the position at t=2.00 s

Use the formula for position under constant acceleration:\[ x = x_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 \]Here, \( x_0 = 0 \), \( v_0 = 0 \):\[ x = 0 + 0 \cdot 2 + \frac{1}{2} \times 1.5625 \times (2.00)^2 = 3.125\, \mathrm{m} \]
04

Determine velocity at t=7.00 s

The puck moves freely from \( t = 2.00 \mathrm{s} \) to \( t = 5.00 \mathrm{s} \) with a constant velocity of \( 3.125\, \mathrm{m/s} \). Then, the same force is applied again for \( t = 5.00 \) to \( t = 7.00 \) (2 seconds). Use the velocity formula again:\[ v = v_0 + a \cdot t \]Where \( v_0 = 3.125 \) at \( t = 5.00 \mathrm{s} \):\[ v = 3.125 + 1.5625 \times 2 = 6.25\, \mathrm{m/s} \]
05

Compute position from t=5.00 to t=7.00 s

To find the total position at \( t = 7.00 \mathrm{s} \), calculate the displacement from \( t = 2.00 \mathrm{s} \) to \( t = 5.00 \mathrm{s} \) and add the displacement from \( t = 5.00 \mathrm{s} \) to \( t = 7.00 \mathrm{s} \).First section (constant velocity):\[ \Delta x_1 = v \cdot t = 3.125 \times 3 = 9.375\, \mathrm{m} \]Second section (acceleration):\[ \Delta x_2 = v_0 \cdot t + \frac{1}{2} a \cdot t^2 = 3.125 \times 2 + \frac{1}{2} \times 1.5625 \times (2)^2 = 8.125\, \mathrm{m} \]Total position:\[ x = 3.125 + 9.375 + 8.125 = 20.625\, \mathrm{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
Constant acceleration is a core concept in Newtonian physics that describes the change in velocity of an object over time when it is subjected to a uniform force. In this example, the hockey puck experiences constant acceleration due to the consistent force applied by the player.
Using Newton's Second Law
  • The formula to find acceleration (\(a\)) is: \[ a = \frac{F}{m} \] where \( F \) is the force and \( m \) is the mass.
  • With a force of 0.250 \(\mathrm{N}\) and a mass of 0.160 \(\mathrm{kg}\), the puck's acceleration is calculated to be 1.5625 \(\mathrm{m/s^2}\).
This constant acceleration is crucial for determining how the puck鈥檚 velocity and position change over time when it slides across a frictionless surface.
Velocity Calculation
Velocity is a vector quantity that refers to the rate at which an object changes its position. To calculate the change in velocity of the puck due to a constant force:
  • Start with the initial velocity \( v_0 = 0 \) since the puck is initially at rest.
  • Apply the formula for velocity under constant acceleration:\[ v = v_0 + a \cdot t \]
For the time interval of 2 seconds, the puck's velocity reaches 3.125 \(\mathrm{m/s}\).
  • At 5 seconds, the force is reapplied for another 2 seconds, increasing the velocity to 6.25 \(\mathrm{m/s}\) by using the same velocity formula.
Velocity calculation helps predict how fast and in what direction the puck will move.
Position Calculation
Calculating the position of an object under constant acceleration involves determining how far it has traveled over a specific time. This is often done using the position-time equation:
  • Given: \( x = x_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 \)
Knowing the puck starts at the origin, at 2 seconds, it moves to 3.125 \(m\). This further includes:
  • The displacement of the puck between 2 and 5 seconds when it's moving at constant velocity (9.375 \(m\)).
  • Lastly, during the second force application from 5 to 7 seconds, the puck moves another 8.125 \(m\), leading to a full position of 20.625 \(m\) at 7 seconds.
Position calculation allows us to track exactly where the puck is on the rink at any given time.
Force and Motion
Force and motion are intrinsically linked in physics, governed by Newton's Second Law. When an unbalanced force acts on an object, it results in acceleration and thus motion.
In this situation, the force applied on the hockey puck sets it into motion, and two important steps occur:
  • The applied force is 0.250 \(N\), creating acceleration (1.5625 \(\mathrm{m/s^2}\)) while the force acts (0-2 seconds and again from 5-7 seconds).
  • Between these times (2-5 seconds), the puck continues moving at constant speed due to its inertia, showcasing another aspect of motion when forces stop acting.
Understanding how force contributes to the motion of an object can help predict changes in its state of rest or uniform movement.
Kinematics
Kinematics is the branch of physics that describes the motion of points, bodies, and systems without considering the forces that cause them. In this exercise, the kinematics of the hockey puck is described using equations of motion.
These kinematic equations allow us to summarize motion in terms of:
  • Velocity: How fast the puck changes position.
  • Acceleration: How quickly the velocity changes.
  • Position: The location of the puck at any given time.
By solving kinematic equations, we can effectively map out the puck's journey on the rink from start to finish under the influence of an external force.
Frictionless Surface
Imagine a perfectly smooth rink where a hockey puck can slide without resistance. A frictionless surface provides this scenario. On such a surface, the only horizontal force acting on the puck is the applied force.
  • Since there is no friction, the force does not need to battle any resistance, allowing the puck to accelerate steadily.
  • When the force is removed, the puck sustains its velocity because there's no opposing friction to gradually slow it down.
Studying motion on a frictionless surface simplifies analysis, illuminating pure effects of applied forces without the complexity of frictional forces.

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Most popular questions from this chapter

A Standing Vertical Jump. Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 \(\mathrm{m}(4 \mathrm{ft}) .\) (This means that he moved upward by 1.2 \(\mathrm{m}\) after his feet left the floor.) Griffith weighed 890 \(\mathrm{N}(200 \mathrm{lb}) .\) (a) What is his speed as he leaves the floor? (b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he was pushing against the floor? (c) Draw his free-body diagram (see Section 4.6\()\) . In terms of the forces on the diagram, what is the net force on him? Use Newton's laws and the results of part (b) to calculate the aver- age force he applied to the ground.

An advertisement claims that a particular automobile can "stop on a dime." What net force would actually be necessary to stop a \(850-\) -kg automobile traveling initially at 45.0 \(\mathrm{km} / \mathrm{h}\) in a distance equal to the diameter of a dime, which is 1.8 \(\mathrm{cm} ?\)

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. The light turns green and you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action- reaction pairs. (The bed of the truck is not friction less.)

A 6.50 -kg instrument is hanging by a vertical wire inside a space ship that is blasting off at the surface of the earth. This ship starts from rest and reaches an altitude of 276 \(\mathrm{m}\) in 15.0 \(\mathrm{s}\) with constant acceleration. (a) Draw a free-body diagram for the instrument during this time. Indicate which force is greater. (b) Find the force that the wire exerts on the instrument.

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a reef at a constant speed of 1.5 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 4.38) .\) When the tanker is 500 \(\mathrm{m}\) from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be sate? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.
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