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Chapter 6: Problem 86

Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very hittle 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 \(\mathrm{s}\) and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

Short Answer

Expert verified
390.0 W

Step by step solution

01

Identify the Known Variables

We know your friend's mass \(m = 65.0 \text{ kg}\), the initial speed \(v_i = 0\text{ m/s}\), the final speed \(v_f = 6.00 \text{ m/s}\), and the time duration of the pull \(t = 3.00 \text{ s}\).
02

Calculate the Acceleration

Use the formula \(a = \frac{v_f - v_i}{t}\) to find the acceleration. Substituting the known values, \(a = \frac{6.00 \text{ m/s} - 0 \text{ m/s}}{3.00 \text{ s}} = 2.00 \text{ m/s}^2\).
03

Determine the Force Applied

Using Newton's second law, \(F = m \cdot a\), calculate the force. With \(m = 65.0 \text{ kg}\) and \(a = 2.00 \text{ m/s}^2\), the force \(F = 65.0 \text{ kg} \times 2.00 \text{ m/s}^2 = 130.0 \text{ N}\).
04

Calculate the Work Done

Work is calculated by \(W = F \cdot d\), where \(d\) is the distance. First, find \(d\) using \(d = v_i \cdot t + \frac{1}{2}a t^2\). Substitute \(v_i = 0\), \(a = 2.00 \text{ m/s}^2\), and \(t = 3.00 \text{ s}\) to get \(d = 0 + \frac{1}{2} \cdot 2.00 \text{ m/s}^2 \cdot (3.00 \text{ s})^2 = 9.00 \text{ m}\). The work \(W = 130.0 \text{ N} \cdot 9.00 \text{ m} = 1170.0 \text{ J}\).
05

Calculate the Average Power

Power is the work done over time, \(P = \frac{W}{t}\). Substitute \(W = 1170.0 \text{ J}\) and \(t = 3.00 \text{ s}\) to get \(P = \frac{1170.0 \text{ J}}{3.00 \text{ s}} = 390.0 \text{ W}\).

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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 one of the fundamental principles in physics. It states that an object will only accelerate when a force is applied to it. This is written mathematically as \( F = m \cdot a \), where
  • \( F \) is the force applied to the object,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration of the object.
In our scenario, your friend is standing on ice, making them nearly frictionless. The ice doesn't apply much resistance, so almost all the force you apply through the rope goes into accelerating your friend. By using Newton's Second Law, we calculate the force needed to reach the desired speed. This equation helps us understand how forces change the motion of objects every day.
Acceleration Calculation
Acceleration is a measure of how quickly an object's velocity changes. It's essential in finding out how fast your friend moved from a standstill during your pull on the icy pond. You determine acceleration with the formula \( a = \frac{v_f - v_i}{t} \), where:
  • \( v_f \) is the final velocity,
  • \( v_i \) is the initial velocity, and
  • \( t \) is the time over which the change occurs.
At first, your friend wasn鈥檛 moving, so \( v_i = 0 \). You helped her reach \( v_f = 6.00 \text{ m/s} \) in \( t = 3.00 \text{ s} \). Plugging in these numbers yields an acceleration of \( 2.00 \text{ m/s}^2 \). Acceleration is straightforward yet critical in predicting how an object鈥檚 speed changes under unbalanced forces like your pulls.
Work and Energy
Work is a physical concept closely tied to force and distance. It tells us how much energy is needed to move an object. In physics, work \( W \) is given by the equation \( W = F \cdot d \), where:
  • \( F \) is the force applied, and
  • \( d \) is the distance over which the force is applied.
Before calculating how much energy it took to get your friend across the ice, we need to determine how far she traveled. This distance is calculated from motion equations like \( d = v_i \cdot t + \frac{1}{2}a t^2 \). Given zero initial velocity, your friend's acceleration over time determines a distance of \( 9.00 \text{ m} \). Using the force from Newton's Second Law, the work you did is \( 1170.0 \text{ J} \). By understanding work, you can grasp how force and movement equate to energy spent.
Average Power Calculation
Power is how fast work gets done or energy gets transferred. It's crucial when assessing the effectiveness of force applied over time. For average power, we use the formula \( P = \frac{W}{t} \), where:
  • \( P \) is the power,
  • \( W \) is the work done, and
  • \( t \) is the time taken.
In the exercise, you completed the work of \( 1170.0 \text{ J} \) in \( 3.00 \text{ s} \). Plugging these values into the formula, you find the average power supplied was \( 390.0 \text{ W} \). This means that the energy transfer to your friend to increase her speed was at a rate of 390 watts. Power gives insight into not just the amount of work done, but how efficiently this work was performed.

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Most popular questions from this chapter

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of \(40.0^{\circ}\) above the horizontal. The glider has mass 0.0900 \(\mathrm{kg}\) . The spring has \(k=640 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 \(\mathrm{m}\) m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 \(\mathrm{m}\) from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{t}-(40 \mathrm{N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\imath}-(3.0 \mathrm{m}) \hat{\mathrm{j}}\) How much work does the force you apply do on the grocery cart?

An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2} .\) (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1}\) , is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2}\) . How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Cycling. For a touring bicyclist the drag coefficient \(C\left(f_{\text { air }}=\frac{1}{2} C A \rho v^{2}\right)\) is \(1.00,\) the frontal area \(A\) is \(0.463 \mathrm{m}^{2},\) and the coefficient of rolling friction is \(0.0045 .\) The rider has mass 50.0 \(\mathrm{kg}\) , and her bike has mass 12.0 \(\mathrm{kg}\) . (a) To maintain a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) (about 27 \(\mathrm{mi} \mathrm{h}\) ) on a level road, what must the rider's power output to the rear wheel be? (b) For racing, the same rider uses a different bike with coefficient of rolling friction 0.0030 and mass 9.00 \(\mathrm{kg}\) . She also crouches down, reducing her drag coefficient to 0.88 and reducing her frontal area to 0.366 \(\mathrm{m}^{2} .\) What must her power output to the rear wheel be then to maintain a speed of 12.0 \(\mathrm{m} / \mathrm{s} ?\) (c) For the situation in part \((\mathrm{b}),\) what power output is required to maintain a speed of 6.0 \(\mathrm{m} / \mathrm{s}\) ? Note the great drop in power requirement when the speed is only halved. (For more on aerodynamic speed limitations for a wide variety of human-powered vehicles, see "The Aerodynamics of Human-Powered Land Vehicles," Scientific American, December 1983.

The mass of a proton is 1836 times the mass of an electron. (a) A proton is traveling at speed \(V\) . At what speed (in terms of \(V )\) would an electron have the same kinetic energy as the proton? (b) An electron has kinetic energy \(K\) . If a proton has the same speed as the electron, what is its kinetic energy (in terms of \(K ) ?\)
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