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Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

Short Answer

Expert verified
The average power is 390 watts.

Step by step solution

01

Analyze the Problem

We need to find the average power supplied by the force applied to your friend on ice. We know the duration of the force application (3.00 seconds), the final velocity (6.00 m/s), and that the initial velocity is zero since your friend starts from rest.
02

Calculate the acceleration

Use the formula for acceleration, which is change in velocity divided by time elapsed: \( a = \frac{v_f - v_i}{t} \). Here, \( v_i = 0 \), \( v_f = 6.00 \ \mathrm{m/s} \), and \( t = 3.00 \ \mathrm{s} \). Substitute the values to find \( a \).
03

Use Newton's Second Law to Find the Force

Apply Newton's second law, \( F = ma \), where \( m = 65.0 \ \mathrm{kg} \) is the mass of your friend. Substitute the calculated acceleration from Step 2 to find the force \( F \).
04

Calculate the Work Done

The work done on your friend can be found using the formula \( W = Fd \), where \( d \) is the distance covered. Use the kinematic equation \( d = v_i t + \frac{1}{2} a t^2 \) to find \( d \), since we know \( v_i = 0 \), \( t = 3.00 \ \mathrm{s} \), and \( a \) from Step 2. Substitute \( d \) and \( F \) to find work \( W \).
05

Calculate the Average Power

Average power is the work done divided by the time: \( P_{avg} = \frac{W}{t} \). Use the work calculated in Step 4 and \( t = 3.00 \ \mathrm{s} \) to find \( P_{avg} \).
06

Final Calculation and Result

Substitute the values obtained from previous steps into the equations to calculate the final average power. Ensure all values are in SI units.

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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 principle that explains how a force affects the motion of an object. This law is expressed by the equation \( F = ma \), which states that the force \( F \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( a \). In the context of pulling your friend on ice, this law directly helps determine the force needed.

In our exercise, the mass of your friend is 65.0 kg. The acceleration, calculated from the change in velocity over time, was determined to be 2.00 m/s². By multiplying these values using Newton's Second Law, we can find the force applied. Remember that force is what causes an object to start moving, stop moving, or change direction.
  • Mass \( m \) = 65.0 kg
  • Acceleration \( a \) = 2.00 m/s²
These calculations show how Newton's Second Law plays a crucial role in determining the necessary force to move your friend from rest to a certain speed on ice.
Kinematics
Kinematics deals with the motion of objects without considering the forces that cause this motion. It provides tools to calculate various parameters of motion such as velocity, displacement, and acceleration.

In this problem, kinematics helps to determine how fast and how far your friend moves after being accelerated. Since she started from rest and accelerated to a velocity of 6.00 m/s over 3.00 seconds, we can use kinematic equations to find these parameters.
First, we establish the velocity change and use it to calculate acceleration. Then, knowing the time and acceleration, we can find the distance moved using the equation:
\[ d = v_i t + \frac{1}{2} a t^2 \]
Where:
  • \( v_i = 0 \) m/s (initial velocity)
  • \( t = 3.00 \) s (time period)
  • \( a = 2.00 \) m/s² (acceleration)
This equation tells us that even when starting from rest, knowing time and acceleration will help find out how far the object travels. Kinematics provides a powerful framework for understanding these types of motion-related problems.
Work-Energy Principle
The work-energy principle connects the concept of work with energy and motion. It states that the work done on an object is equal to the change in its kinetic energy. This principle is essential when calculating average power applied, as in our exercise.

First, we calculate the work done as the product of force and the distance over which it acts: \( W = Fd \). We previously found force using Newton's Second Law and distance using kinematics.

Once work is determined, the average power can be calculated as work divided by the time period in which the force was applied. This gives us insight into the rate at which energy is used or transferred by a force:
\[ P_{avg} = \frac{W}{t} \]
Where:
  • \( W \) is the work done
  • \( t = 3.00 \) s
The work-energy principle shows the intricate relationship between work, energy, and power in dynamic situations, enabling better understanding of how systems interact and consume energy. By mastering this principle, we gain the ability to solve numerous practical real-world physics problems.

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Most popular questions from this chapter

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length.(a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.

CALC An airplane in flight is subject to an air resistance force proportional to the square of its speed \(v .\) But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.104). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to \(v^{2},\) so that the total air resistance force can be expressed by \(F_{\text { air }}=\alpha v^{2}+\beta / v^{2},\) where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna \(150,\) a small single-engine airplane, \(\alpha=0.30 \mathrm{N} \cdot \mathrm{s}^{2} / \mathrm{m}^{2}\) and \(\beta=3.5 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\) In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).

A 4.80 -kg watermelon is dropped from rest from the roof of a 25.0 -m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

CALC An object has several forces acting on it. One of these forces is \(\vec{\boldsymbol{F}}=a x y \hat{\boldsymbol{r}},\) a force in the \(x\) -direction whose magni- tude depends on the position of the object, with \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{2}\) . Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point \(x=0\) , \(y=3.00 \mathrm{m}\) and moves parallel to the \(x\) -axis to the point \(x=2.00 \mathrm{m}, y=3.00 \mathrm{m} .\) (b) The object starts at the point \(x=2.00 \mathrm{m}, \quad y=0\) and moves in the \(y\) -direction to the the point \(x=2.00 \mathrm{m}, y=3.00 \mathrm{m} .\) (c) The object starts at the origin and moves on the line \(y=1.5 x\) to the point \(x=2.00 \mathrm{m},\) \(y=3.00 \mathrm{m} .\)

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