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Chapter 5: Problem 68

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110\) \(\mathrm{m}\) and at an angle of \(34.10^{\circ}\), and it lands at a horizontal distance of \(15.90 \mathrm{~m}\). What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

Short Answer

Expert verified
The average force exerted by the athlete is approximately 367.38 N.

Step by step solution

01

Analyze the Problem

We need to determine the average force exerted by the athlete on the shot put during the acceleration phase along a ramp, using given information about initial and final velocities, distance, and angle.
02

Resolve Velocities

First, calculate the final velocity of the shot using the projectile motion formula for horizontal range:\[R = \frac{v^2 \sin(2\theta)}{g}\]where \(R = 15.90 \, \mathrm{m}\) and \(\theta = 34.10^\circ\). Solving for \(v\), the launch speed.
03

Calculate Final Speed

Rearranging the range equation:\[v^2 = \frac{R \cdot g}{\sin(2\theta)}\]Substitute the values:\[v^2 = \frac{15.90 \, \mathrm{m} \cdot 9.81 \, \mathrm{m/s^2}}{\sin(68.2^\circ)}\]\[v \approx 12.54 \, \mathrm{m/s}\]
04

Apply Work-Energy Principle

The work-energy principle states that the work done by the force is equal to the change in kinetic energy:\[W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)\]where \(v_i = 2.500 \, \mathrm{m/s}\) is the initial speed and \(v_f = 12.54 \, \mathrm{m/s}\) is the final speed.
05

Calculate Change in Kinetic Energy

Calculate \(\Delta KE\):\[\Delta KE = \frac{1}{2} \times 7.260 \, \mathrm{kg} \times (12.54^2 - 2.500^2) \, \mathrm{m/s}^2\]\[\Delta KE \approx 533.96 \, \mathrm{J}\]
06

Calculate Force from Work

The work done is also equal to the force times the distance along the direction of the force, considering the angle:\[W = F \cdot d \cdot \cos(\theta)\]Solving for \(F\), with \(d = 1.650 \, \mathrm{m}\) and \(\theta = 34.10^\circ\):\[F = \frac{\Delta KE}{d \cdot \cos(\theta)}\]\[F \approx \frac{533.96 \, \mathrm{J}}{1.650 \, \mathrm{m} \times \cos(34.10^\circ)}\]\[\approx 367.38 \, \mathrm{N}\]
07

Conclusion

The magnitude of the athlete's average force on the shot during the acceleration phase is approximately \(367.38 \, \mathrm{N}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept in physics and particularly in the study of projectile motion. It describes the energy possessed by an object due to its motion. The equation for kinetic energy (KE) is given by:\[KE = \frac{1}{2} mv^2\]where \(m\) is the mass of the object and \(v\) is its velocity. In the context of this exercise, we initially calculate the change in kinetic energy as the shot put accelerates from its initial velocity to its final velocity.
This change is crucial because it tells us how much energy has been gained by the shot put due to the athlete's action. When we computed the change in kinetic energy for this problem, it helped us determine how much work was done on the shot. Understanding kinetic energy helps us to bridge how motion relates to the forces applied to the shot put.
Work-Energy Principle
The work-energy principle is a powerful tool used to relate the work done on an object to its change in kinetic energy. This principle states:\[W = \Delta KE\]This implies that the work done (\(W\)) on an object equals the change in its kinetic energy (\(\Delta KE\)). In the given scenario, this principle is leveraged to calculate the work done by the athlete.
  • Initial kinetic energy was computed based on the shot's initial speed.
  • Final kinetic energy was derived from the final speed after acceleration.
  • The difference gave us the change in kinetic energy, \(\Delta KE\).
By applying the work-energy principle, we converted the idea of work into a practical calculation involving force, distance, and the angle of motion, which eventually let us find the magnitude of the average force.
Horizontal Range
The horizontal range of a projectile is a major aspect when analyzing motion arcs. It refers to the horizontal distance a projectile travels during its flight. The formula to calculate the horizontal range \(R\) is:\[R = \frac{v^2 \sin(2\theta)}{g}\]Here, \(v\) is the launch speed, \(\theta\) the launch angle, and \(g\) the acceleration due to gravity.
For the shot put problem, this formula was essential in determining the final launch speed required for the shot to cover the specified horizontal distance of 15.90 meters. Understanding this equation not only helped in finding critical velocities but also illustrated how initial conditions like speed and angle significantly influence projectile motion outcomes, such as the range achieved in this exercise.
Average Force
Average force is a key concept for understanding how much an athlete influences the motion of a shot put during the acceleration phase. The force can be related to work and energy using the equation:\[W = F \cdot d \cdot \cos(\theta)\]where \(F\) is the force, \(d\) the distance covered during the force application, and \(\theta\) the angle of the force relative to the direction of motion.
In our exercise, the average force exerted by the athlete was calculated by substituting the already found work (from the change in kinetic energy) and adjusting for the angle of the ramp-like path. Calculating this force is paramount, as it quantifies the effect of the athlete's push against the resistance faced due to factors like gravity and the shot's weight, eventually leading to the targeted range.

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Most popular questions from this chapter

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

While two forces act on it, a particle is to move at the constant velocity \(\vec{v}=(3 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(4 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) One of the forces is \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+(-6 \mathrm{~N}) \hat{\mathrm{j}}\) What is the other force?

You pull a short refrigerator with a constant force \(\vec{F}\) across a greased (frictionless) floor, either with \(\vec{F}\) horizontal (case 1\()\) or with \(\vec{F}\) tilted upward at an angle \(\theta\) (case 2). (a) What is the ratio of the refrigerator's speed in case 2 to its speed in case 1 if you pull for a certain time \(t ?\) (b) What is this ratio if you pull for a certain distance \(d ?\)

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 N) \hat{j}+(-2 N) \hat{k}\) and \(\vec{F}_{2}=(-5 N) \hat{i}+(8 N) \hat{j}+(-2 N) \hat{k} .\) What is the third force?
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