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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; she 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
The position and speed of the puck at \(t=2.00 s\) are \(3.125 m\) and \(3.125 m/s\) respectively. The position and speed of the puck at \(t=7.00 s\) are \(12.5 m\) and \(6.25 m/s\) respectively.

Step by step solution

01

Calculate Acceleration

The applied force on the puck is given and the mass of the puck is also provided, which allows the calculation of the acceleration using Newton's second law. The acceleration \(a\) is given by the quotient of the force \(F\) and the mass \(m\): \(a = F / m\). Substituting the given values, \(a = 0.250 N / 0.160 kg = 1.5625 m/s^2.\)
02

Use Kinematic Equations to Find Position & Speed

The position and speed of the puck at \(t=2.00 s\) can be obtained using the kinematic equations. In this case, the initial speed \(u\) is 0, the acceleration \(a\) is \(1.5625 m/s^2\), and the time \(t\) is \(2.00 s\). The final speed \(v\) can be calculated with \(v = u + at\), which gives \(v = 0 + 1.5625 m/s^2 * 2.00 s = 3.125 m/s\). The position \(s\) can be calculated with \(s = ut + 0.5*at^2\), which gives \(s = 0 + 0.5*1.5625 m/s^2 * (2.00 s)^2 = 3.125 m\)
03

Calculate Position & Speed at \(t=7.00 s\)

The force again is applied at \(t=5.00 s\), so the puck will again get accelerated for 2 seconds. Taking the final speed obtained in previous step as the initial speed, we can again use the kinematic equations to find the speed and displacement at \(t=7.00 s\). The final speed then becomes \(v = 3.125 m/s + 1.5625 m/s^2 * 2.00 s = 6.25 m/s\). The displacement during these 2 seconds is \(s = 3.125 m *2.00 s + 0.5*1.5625 m/s^2 * (2.00 s)^2 = 9.375 m\). The total displacement from the origin at \(t=7.00 s\) is thus the sum of the previous position and the displacement now, i.e., \(3.125 m + 9.375 m = 12.5 m\).

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Key Concepts

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

Newton's second law
Newton's second law is a fundamental concept in physics, especially crucial in understanding motion. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be summarized in the equation:\[ a = \frac{F}{m} \]where:
  • {\(a\)} is the acceleration,
  • {\(F\)} is the net force acting on the object,
  • {\(m\)} is the mass of the object.
This law is essential because it helps us predict how an object will move when different forces are applied. For instance, in the case of the hockey puck, knowing the force applied and the mass allows us to calculate the acceleration. The player pushes with a force of {0.250 N}, and with the known mass of {0.160 kg}, we find that the puck's acceleration is {1.5625 m/s²}. This shows how the force-to-mass ratio determines the acceleration.
Kinematic equations
Kinematic equations are vital tools in physics that help describe the motion of objects without considering the forces behind them. They are used to calculate an object's position, velocity, and time given its acceleration. These equations are especially useful when acceleration is constant.
The basic kinematic equations are:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
where:
  • {\(v\)} is the final velocity,
  • {\(u\)} is the initial velocity,
  • {\(a\)} is the acceleration,
  • {\(s\)} is the displacement,
  • {\(t\)} is the time.
In our exercise, these equations help us determine the speed and position at different times. At {t=2.00 s}, since the puck started from rest (\(u = 0\)), we use these equations to find that both the final velocity and position are {3.125 m/s} and {3.125 m}, respectively. This demonstrates how knowing just a few initial conditions can predict future motion.
Acceleration calculation
Acceleration is a measure of how quickly an object's velocity changes. It's crucial for understanding motion. We often calculate acceleration using Newton's second law, especially when an object is subject to a constant force.
In the hockey puck exercise, calculating the acceleration is the first step in understanding how the puck moves. With a force of {0.250 N} and a mass of {0.160 kg}, the acceleration simplifies to {1.5625 m/s²}. This is a steady increase in speed over time.
Understanding acceleration is essential because it affects both the speed and position of the puck. After the force stops, if the surface were not frictionless, other forces like friction would alter this acceleration. However, on a frictionless surface like an ice rink, the puck maintains a constant velocity after the force is removed, making the initial acceleration calculation even more significant in predicting its future motion.

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