/*! 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 85 A water-skier is being pulled by... [FREE SOLUTION] | 91Ó°ÊÓ

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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) What type of energy is changing? (b) Is the work being done by the net external force acting on the skier positive, zero, or negative? Why? (c) How is this work related to the change in the energy of the skier? 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 kinetic energy of the skier is increasing. Positive work is done by the external force, increasing the skier's kinetic energy. Work done is equal to the change in kinetic energy: approximately 2339 J.

Step by step solution

01

Identify the situation

The skier is being pulled by the boat, which means there is an external force acting on the skier, causing acceleration. As a result, the speed of the skier is increasing.
02

Determine type of energy change

When a skier accelerates, the kinetic energy is changing because the speed of the skier is increasing. Kinetic energy is directly related to the speed of the object.
03

Assess work done by external force

Since the skier accelerates and gains speed, the work done by the net external force (the force from the rope) must be positive. Positive work transfers energy to the skier, increasing kinetic energy.
04

Relation between work and energy

The work done on the skier by the net external force is equal to the change in kinetic energy of the skier. This follows from the work-energy principle: Work done = Change in kinetic energy.
05

Calculate initial and final kinetic energy

The initial kinetic energy (KE_i) is calculated as \( KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2} \times 70.3 \times 6.1^2 \). The final kinetic energy (KE_f) is calculated as \( KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2} \times 70.3 \times 11.3^2 \).
06

Calculate the change in kinetic energy

The change in kinetic energy is \( \Delta KE = KE_f - KE_i \). Substitute the values from Step 5 to find \( \Delta KE \).
07

Conclusion

The work done by the net external force is equal to the change in kinetic energy. Substitute the kinetic energy values from Step 6 to find the work done.

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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 the energy that an object possesses due to its motion. It's a fundamental concept in physics and is pivotal in understanding how objects move. The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. This means that the energy is directly proportional to both the mass of the object and the square of its speed.
Since kinetic energy depends on speed, any change in the velocity of an object—such as a water-skier speeding up—results in a change in kinetic energy.
This is exactly what happens when the skier is pulled faster by the boat. His speed increases from 6.10 m/s to 11.3 m/s, leading to an increase in kinetic energy.
  • Initial kinetic energy: calculated using the initial speed.
  • Final kinetic energy: calculated with the new, higher speed.
Therefore, understanding kinetic energy is crucial for analyzing how the skier's motion changes.
External Force
An external force refers to the force applied to an object by something in its environment. In the case of our water-skier, the tow rope connected to the boat is the source of the external force. This force is crucial because it causes the skier to accelerate, changing his speed.
External forces can do work on an object, adding or subtracting energy. In this scenario, the rope pulls the skier, transferring energy from the boat's engine to the skier.
Here's what happens:
  • The force pulls the skier at an angle or along a straight path.
  • As the skier moves, this force does work, translating energy from the boat to the skier, increasing his speed.
By understanding the external force acting on the skier, we can deduce that positive work is done, increasing his kinetic energy.
Acceleration
Acceleration is the rate at which an object's velocity changes. It can be due to a change in speed or direction. For a water-skier, acceleration happens when the boat increases its throttle, speeding up the skier.
The link between force and acceleration is expressed by Newton's Second Law, \( F = ma \). This law implies that the external force from the tow rope causes the skier's mass to accelerate.
With acceleration, the skier's speed changes over time:
  • If acceleration is in the direction of motion, the skier gains speed.
  • If it's against the direction, the skier slows down.
Since the skier's speed increased—from 6.10 m/s to 11.3 m/s—it indicates that the acceleration was positive. This ties back to the work-energy principle, as increasing speed leads to more kinetic energy.
Energy Change
Energy change occurs when work is done on or by an object leading to a shift in energy. In our water-skiing scenario, we witness energy change primarily as a change in kinetic energy.
When the external force (tow rope) does work by accelerating the skier, this work correlates directly to an energy change as per the work-energy principle.
This principle states:
The work done on an object equals the change in its energy.
  • Initial kinetic energy: when speed was 6.10 m/s.
  • Final kinetic energy: when speed rose to 11.3 m/s.
The difference between these kinetic energy values (final - initial) represents the energy change due to work done.
This explanation makes it clear that as the skier was pulled faster, energy was transferred to him, enhancing his motion.

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Most popular questions from this chapter

Two cars, \(A\) and \(B\), are traveling with the same speed of \(40.0 \mathrm{~m} / \mathrm{s}\), each having started from rest. Car A has a mass of \(1.20 \times 10^{3} \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(2.00 \times 10^{3} \mathrm{~kg} .\) Compared to the work required to bring car A up to speed, how much additional work is required to bring car B up to speed?

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 a distance of \(1.9 \mathrm{~m}\) up the slope, the speed of the skier is \(4.4 \mathrm{~m} / \mathrm{s}\). (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 6200 -kg satellite is in a circular earth orbit that has a radius of \(3.3 \times 10^{7} \mathrm{~m}\). A net external force must act on the satellite to make it change to a circular orbit that has a radius of \(7.0 \times 10^{6} \mathrm{~m}\). What work must the net external force do?

A \(2.40 \times 10^{2}-N\) force is pulling an \(85.0\) -kg refrigerator across a horizontal surface. The force acts at an angle of \(20.0^{\circ}\) above the surface. The coefficient of kinetic friction is \(0.200\), and the refrigerator moves a distance of \(8.00 \mathrm{~m}\). Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

The concepts in this problem are similar to those in Multiple-Concept Example \(4,\) except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 79 -kg person straight up by means of a cable. The person has an upward acceleration of \(0.70 \mathrm{~m} / \mathrm{s}^{2}\) and is lifted from rest through a distance of \(11 \mathrm{~m}\). (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work- energy theorem and find the final speed of the person.

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