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Chapter 9: Problem 59

A Sprinter A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s} ?\) (c) What average power does the sprinter generate during the 1.6 s interval?

Short Answer

Expert verified
Speed = 14.0 m/s, Kinetic Energy = 6700 J, Average Power = 4187.5 W.

Step by step solution

01

Understand the Problem

A sprinter who weighs 670 N runs the first 7.0 m in 1.6 s, starting from rest and accelerating uniformly. Need to find (a) speed, (b) kinetic energy, and (c) average power generated during the 1.6s interval.
02

Calculate Acceleration

Use the equation of motion for uniform acceleration: \[ s = ut + \frac{1}{2}at^2 \] Given, initial speed (u) = 0, s = 7.0 m, t = 1.6 s. Substitute the values to find acceleration (a).\[ 7.0 = 0 + \frac{1}{2}a(1.6^2) \]\[ 7.0 = 0.8a \]\[ a = \frac{7.0}{0.8} \]\[ a = 8.75 \text{ m/s}^2 \]
03

Calculate Final Speed

To find the final speed (v), use the equation: \[ v = u + at \]Given u = 0, a = 8.75 m/s虏, t = 1.6 s.\[ v = 0 + 8.75 \times 1.6 \]\[ v = 14.0 \text{ m/s} \]
04

Calculate Mass

Weight (W) is given by: \[ W = mg \]Given W = 670 N, g = 9.8 m/s虏. Solve for mass (m).\[ m = \frac{670}{9.8} \]\[ m = 68.4 \text{ kg} \]
05

Calculate Kinetic Energy

Use the formula for kinetic energy (KE): \[ KE = \frac{1}{2}mv^2 \]Given m = 68.4 kg, v = 14.0 m/s.\[ KE = \frac{1}{2} \times 68.4 \times (14.0)^2 \]\[ KE = 1/2 \times 68.4 \times 196 \]\[ KE = 6700 \text{ J} \]
06

Calculate Average Power

Average power (P) is given by: \[ P = \frac{\text{Work}}{t} \]Work done is equal to the change in kinetic energy.Given KE = 6700 J, t = 1.6 s. \[ P = \frac{6700}{1.6} \]\[ P = 4187.5 \text{ W} \]

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to a constant change in velocity over a given period of time. It means that the acceleration, which is the rate of change of velocity, does not vary. When you move with uniform acceleration, your speed increases by the same amount each second.
This idea can be captured using the equation of motion: \[ s = ut + 0.5at^2 \]Where:
  • \( s \): distance traveled
  • \( u \): initial speed
  • \( t \): time taken
  • \( a \): acceleration
In this exercise, the sprinter starts from rest, so \( u \) is zero. By inserting the given values into the formula, we can determine the uniform acceleration as follows:\[ 7.0 = 0.5 \times a \times (1.6^2) \]which simplifies to \[ a = 8.75 \text{ m/s}^2 \] This shows the sprinter's change in velocity per second squared during the first 1.6 seconds.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. The formula to calculate kinetic energy is:\[ KE = 0.5 \times m \times v^2 \]Where:
  • \( KE \): kinetic energy
  • \( m \): mass of the object
  • \( v \): velocity of the object
In the exercise, after 1.6 seconds, the sprinter reached a velocity of 14.0 m/s. Given the sprinter鈥檚 weight was 670 N, we calculated his mass as 68.4 kg.Using these values, the kinetic energy at 1.6 seconds is:\[ KE = 0.5 \times 68.4 \times (14.0)^2 \]\[ KE = 6700 \text{ J} \] Therefore, at the end of 1.6 seconds, the sprinter's kinetic energy is 6700 joules.
Average Power
Average power is the amount of work done per unit of time. It鈥檚 a measure of how quickly energy is being used. The formula for average power is:\[ P = \frac{ \text{Work} }{ t } \] In the exercise, the sprinter鈥檚 work done is equivalent to the change in kinetic energy. We calculated this change in kinetic energy as 6700 J. Given that this occurs over 1.6 seconds, we find the average power as:\[ P = \frac{ 6700 }{ 1.6 } \]\[ P = 4187.5 \text{ W} \] So, the average power generated by the sprinter during the first 1.6 seconds is 4187.5 watts. This demonstrates the high rate of energy conversion required to accelerate from rest to 14.0 m/s in such a short time.

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Most popular questions from this chapter

Velodrome (a) In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN}\), was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) \(\operatorname{In}\) 1960, Mrs. Maxwell Rogers of Tampa, Florida, reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm}\), how much work did her force do on the car?

Transporting Boxes Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At a certain location the conveyor belt moves for \(2.0 \mathrm{~m}\) up an incline that makes an angle of \(10^{\circ}\) with the horizontal, then for \(2.0 \mathrm{~m}\) horizontally, and finally for \(2.0 \mathrm{~m}\) down an incline that makes an angle of \(10^{\circ}\) with the horizontal. Assume that a \(2.0 \mathrm{~kg}\) box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box (a) as the box moves up the \(10^{\circ}\) incline, (b) as the box moves horizontally, and (c) as the box moves down the \(10^{\circ}\) incline?

Coin on a Frictionless Plane A coin slides over a frictionless plane and across an \(x y\) coordinate system from the origin to a point with \(x y\) coordinates \((3.0 \mathrm{~m}, 4.0 \mathrm{~m})\) while a constant force acts on it. The force has magnitude \(2.0 \mathrm{~N}\) and is directed at a counterclockwise angle of \(100^{\circ}\) from the positive direction of the \(\bar{x}\) axis. How much work is done by the force on the coin during the displacement?

Truck Traveling North A \(2100 \mathrm{~kg}\) truck traveling north at \(41 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(51 \mathrm{~km} / \mathrm{h}\). (a) What is the change in the kinetic energy of the truck? What are the (b) magnitude and (c) direction of the change in the translational momentum of the truck?

Freight Car A railroad freight car of mass \(3.18 \times 10^{4} \mathrm{~kg}\) collides with a stationary caboose car. They couple together, and \(27.0 \%\) of the initial kinetic energy is transferred to nonconservative forms of energy (thermal, sound, vibrational, and so on). Find the mass of the caboose. Hint: Translational momentum is conserved.
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