/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Go A water-skier is being pulled... [FREE SOLUTION] | 91Ó°ÊÓ

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Go A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 70.3 -kg water- skier has an initial speed of \(6.10 \mathrm{m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{m} / \mathrm{s} .\) Determine the work done by the net external force acting on the skier.

Short Answer

Expert verified
The work done is 3183.8 J.

Step by step solution

01

Understand the Problem

We need to find the work done on the skier given her change in speed. To do this, we use the concept of kinetic energy and the work-energy principle.
02

Apply the Work-Energy Principle

The work-energy principle states that the work done by the net external force on an object is equal to the change in kinetic energy of the object. Mathematically, this can be written as:\[ W = \Delta KE = KE_f - KE_i \]where \( KE_i \) is the initial kinetic energy and \( KE_f \) is the final kinetic energy.
03

Calculate Initial Kinetic Energy

The initial kinetic energy \( KE_i \) can be calculated using the formula:\[ KE_i = \frac{1}{2} m v_i^2 \]Substitute \( m = 70.3 \text{ kg} \) and \( v_i = 6.10 \text{ m/s} \) to find:\[ KE_i = \frac{1}{2} \times 70.3 \times (6.10)^2 = 1304.835 \text{ J} \]
04

Calculate Final Kinetic Energy

The final kinetic energy \( KE_f \) is calculated similarly using the final velocity:\[ KE_f = \frac{1}{2} m v_f^2 \]Substitute \( m = 70.3 \text{ kg} \) and \( v_f = 11.3 \text{ m/s} \) to find:\[ KE_f = \frac{1}{2} \times 70.3 \times (11.3)^2 = 4488.635 \text{ J} \]
05

Calculate the Work Done

Now, calculate the work done \( W \) using the change in kinetic energy:\[ W = KE_f - KE_i = 4488.635 - 1304.835 = 3183.8 \text{ J} \]
06

Final Step: Conclusion

The work done by the net external force acting on the skier is \( 3183.8 \text{ J} \).

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a crucial concept in physics that refers to the energy that an object possesses due to its motion. The formula to calculate kinetic energy (KE) is: \[ KE = \frac{1}{2} mv^2 \] where:
  • \(m\) is the mass of the object in kilograms,
  • \(v\) is the velocity of the object in meters per second.
In our exercise, the water-skier has both initial and final speeds. Therefore, we need to calculate both the initial kinetic energy \(KE_i\) and the final kinetic energy \(KE_f\). This calculation helps in understanding how much energy is gained or lost by the skier as her speed changes. By substituting the skier's mass and velocities into the kinetic energy formula, we first find the initial state kinetic energy. Then, we do the same for the final state to understand how the skier's energy has transformed. These are preliminary calculations before using the work-energy principle.
The Concept of Work Done
Work done is another foundational idea in physics, directly connected to energy changes. In this context, work done refers to the energy transferred by the force to cause the skier's acceleration. The work done by a force is given by the equation: \[ W = F \cdot d \cdot \cos(\theta) \] where:
  • \(F\) is the force applied,
  • \(d\) is the distance over which the force is applied,
  • \(\theta\) is the angle between the force and displacement directions.
However, when using the work-energy principle, we focus on the change in kinetic energy to determine the work done. For the skier, this can be found using: \[ W = KE_f - KE_i \] This formula indicates that the work done on the skier by the net external force is essentially the difference in her kinetic energy from start to finish. It translates the forces acting upon her into a measurable energy change, which explains her acceleration.
Role of Net External Force
Net external force is the key player in changing the state of motion of an object and is defined as the overall force resulting from all the external forces acting on an object. In physics, it's what causes acceleration, which is famously stated in Newton's second law: \[ F_{net} = ma \] where:
  • \(F_{net}\) is the net external force,
  • \(m\) is the mass of the object,
  • \(a\) is the acceleration.
In the water-skier scenario, the net external force results from the boat's pull through the tow rope. This force is responsible for the skier's acceleration from an initial speed to a higher speed. By calculating the work done using the change in kinetic energy, we're effectively determining how much energy the net external force contributes to make the skier go faster. Understanding the net external force not only helps us compute work but also deepens our grasp of the dynamics involved in motion and energy transfer in real-world situations.

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Most popular questions from this chapter

A 63-kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{m} / \mathrm{s}\). After coasting \(1.9 \mathrm{m}\) up the slope, the skier has a speed of \(4.4 \mathrm{m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{m} / \mathrm{s} .\) The building is \(31.0 \mathrm{m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

The (nonconservative) force propelling a \(1.50 \times 10^{3}-\mathrm{kg}\) car up a mountain road does \(4.70 \times 10^{6} \mathrm{J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta .\) One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0}\). On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

The hammer throw is a track-and-field event in which a \(7.3 \mathrm{kg}\) ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of \(29 \mathrm{m} / \mathrm{s} .\) For comparison, a .22 caliber bullet has a mass of \(2.6 \mathrm{g}\) and, starting from rest, exits the barrel of a gun at a speed of \(410 \mathrm{m} / \mathrm{s}\). Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

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