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Chapter 5: Problem 11

A \(65.0-\mathrm{kg}\) runner has a speed of \(5.20 \mathrm{~m} / \mathrm{s}\) at one instant during a long-distance event. (a) What is the runner's kinetic energy at this instant? (b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?

Short Answer

Expert verified
The initial kinetic energy of the runner is \(881.00 \, \mathrm{J}\). When the runner doubles his speed, his kinetic energy increases by a factor of \(4\).

Step by step solution

01

Calculation of the initial kinetic energy

Insert the given values into the kinetic energy formula \(KE = \frac{1}{2} m v^2\). In this case, \(m = 65.0 \, \mathrm{kg}\) and \(v = 5.20 \, \mathrm{m/s}\). Calculate the value.
02

Calculation of the kinetic energy after speed doubles

To find the new kinetic energy, replace \(v\) in the formula with the doubled speed value. That is, \(v = 2 \times 5.20 \, \mathrm{m/s}\). Calculate the new kinetic energy.
03

Calculate the ratio of the kinetic energy change.

To find the factor by which the kinetic energy changes when the speed is doubled, divide the new kinetic energy by the initial kinetic energy.

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Key Concepts

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

Physics Problem Solving
In physics, problem solving is a methodical approach that involves breaking down a problem into manageable steps to reach a solution.
When dealing with problems like calculating kinetic energy, two main steps are essential: analyzing given data and applying the correct formulas. In our example, the kinetic energy quantity relates to a runner with a known mass and speed.
By understanding the conditions, one can choose the kinetic energy equation, which forms the conceptual basis for solving this problem.
  • Step 1 involved identifying the runner鈥檚 mass and speed within the kinetic energy formula: \( KE = \frac{1}{2} m v^2 \).
  • Step 2 required using the identified formula to calculate the actual kinetic energy value after doubling the runner鈥檚 speed.
  • Step 3 entailed calculating the change in kinetic energy due to the variation in speed.
Through careful analysis and substitution, problem-solving outlines how physical principles can predict outcomes, such as the effect of speed changes on kinetic energy.
Work and Energy
Work and energy are closely related concepts in physics that describe how forces affect objects and their motion.
Kinetic energy, a type of energy related to motion, quantifies how much work an object can perform based on its movement.
For the runner's scenario, kinetic energy depends significantly on speed. The kinetic energy formula, \( KE = \frac{1}{2} m v^2 \), shows that kinetic energy is proportional to the square of speed.
  • When the runner鈥檚 speed doubled, calculate the kinetic energy using the new speed value. This shows energy's strong dependence on velocity.
  • As speed increases, kinetic energy rises exponentially, not linearly, highlighting the physical significance of movement in calculating energy.
  • The exercise demonstrates the larger energy required for higher speeds, visualizing energy concepts in real-life situations like a runner's pace during a race.
Understanding how energy increases with speed changes allows learners to grasp why energy consumption is critical in athletic performance.
Speed and Velocity
Speed and velocity are fundamental elements in physics, defining how fast and in what direction an object moves.
While speed is a scalar quantity鈥攊ndicating how fast an object moves without direction鈥攙elocity includes both speed and direction, making it a vector quantity.
In our exercise, the runner's speed is the primary factor influencing kinetic energy.
  • Initial velocity was \(5.20 \text{ m/s}\), forming the basis of the kinetic energy calculation.
  • Doubling this speed to \(10.40 \text{ m/s}\) changed the kinetic energy proportionally due to its squared relationship in the equation.
  • The variation illustrated how changes in speed, even without altering direction, substantially impact energy calculations.
Recognizing the difference between speed and velocity helps in forecasting object's motion effects, particularly useful in understanding real-world problems like optimizing athletic performance and efficiency.

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Most popular questions from this chapter

Hooke's law describes a certain light spring of unstretched length \(35.0 \mathrm{~cm}\). When one end is attached to the top of a door frame and a \(7.50-\mathrm{kg}\) object is hung from the other end, the length of the spring is \(41.5 \mathrm{~cm}\). (a) Find its spring constant. (b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of \(190 \mathrm{~N}\). Find the length of the spring in this situation.

A roller-coaster car of mass \(1.50 \times 10^{3} \mathrm{~kg}\) is initially at the top of a rise at point (A). It then moves \(35.0 \mathrm{~m}\) at an angle of \(50.0^{\circ}\) below the horizontal to a lower point (B). (a) Find both the potential energy of the system when the car is at points (A) and (B) and the change in potential energy as the car moves from point (A) to point (B), assuming \(y=0\) at point. (B). (b) Repeat part (a), this time choosing \(y=0\) at point (C), which is another \(15.0 \mathrm{~m}\) down the same slope from point (B).

A \(3.50-\mathrm{kN}\) piano is lifted by three workers at constant speed to an apartment \(25.0 \mathrm{~m}\) above the street using a pulley system fastened to the roof of the building. Each worker is able to deliver 165 W of power, and the pulley system is \(75.0 \%\) efficient (so that \(25.0 \%\) of the mechanical energy is lost due to friction in the pulley). Neglecting the mass of the pulley, find the time required to lift the piano from the street to the apartment.

W A child and a sled with a combined mass of \(50.0 \mathrm{~kg}\) slide down a frictionless slope. If the sled starts from rest and has a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) at the bottom, what is the height of the hill?

GP (a) A block with a mass \(m\) is pulled along a horizontal surface for a distance \(x\) by a constant force \(\overrightarrow{\mathbf{F}}\) at an angle \(\theta\) with respect to the horizontal. The coefficient of kinetic friction between block and table is \(\mu_{k}\). Is the force exerted by friction equal to \(\mu_{k}\) mg? If not, what is the force exerted by friction? (b) How much work is done by the friction force and by \(\overrightarrow{\mathbf{F}}\) ? (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let \(m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}\), \(F=15.0 \mathrm{~N}\), and \(\mu_{k}=0.400\), and find the answers to parts (a) and (b).
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