**Problem Statement:**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 79.5-kg water-skier has an initial speed of 7.4 m/s. Later, the speed increases to 10.6 m/s. Determine the work done by the net external force acting on the skier.---To solve this problem, you can use the work-energy principle which states:\[ \text{Work done} = \Delta K.E. \]where \( \Delta K.E. \) (change in kinetic energy) is given by:\[ \Delta K.E. = \frac{1}{2} m(v_f^2 - v_i^2) \]Here:- \( m \) = mass of skier = 79.5 kg- \( v_i \) = initial speed = 7.4 m/s- \( v_f \) = final speed = 10.6 m/sNow, substituting the values, you get:\[ \Delta K.E. = \frac{1}{2} \times 79.5 \times (10.6^2 - 7.4^2) \]Calculate the squares of the speeds:\[ 10.6^2 = 112.36 \]\[ 7.4^2 = 54.76 \]Now, subtract:\[ 112.36 - 54.76 = 57.6 \]Finally,\[ \Delta K.E. = \frac{1}{2} \times 79.5 \times 57.6 \]\[ \Delta K.E. = 39.75 \times 57.6 \]\[ \Delta K.E. = 2289.6 \text{ J} \]Therefore, the work done by the net external force acting on the skier is 2289.6 Joules.

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Description: **Problem Statement:**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 79.5-kg water-skier has an initial speed of 7.4 m/s. Later, the speed increases to 10.6 m/s. De

Mass (m) = 79.5 kgInitial velocity (u) = 7.4 m/sFinal velocity (v) = 10.6 m/sRequired:Work done (W)

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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 79.5-kg water-skier has an initial speed of 7.4 m/s. Later, the speed increases to 10.6 m/s. Determine the work done by the net external force acting on the skier.

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 79.5-kg water-skier has an initial speed of 7.4 m/s. Later, the speed increases to 10.6 m/s. Determine the work done by the net external force acting on the skier.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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**Problem Statement:**

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 79.5-kg water-skier has an initial speed of 7.4 m/s. Later, the speed increases to 10.6 m/s. Determine the work done by the net external force acting on the skier.

---

To solve this problem, you can use the work-energy principle which states:

\[ \text{Work done} = \Delta K.E. \]

where \( \Delta K.E. \) (change in kinetic energy) is given by:

\[ \Delta K.E. = \frac{1}{2} m(v_f^2 - v_i^2) \]

Here:

- \( m \) = mass of skier = 79.5 kg
- \( v_i \) = initial speed = 7.4 m/s
- \( v_f \) = final speed = 10.6 m/s

Now, substituting the values, you get:

\[ \Delta K.E. = \frac{1}{2} \times 79.5 \times (10.6^2 - 7.4^2) \]

Calculate the squares of the speeds:

\[ 10.6^2 = 112.36 \]
\[ 7.4^2 = 54.76 \]

Now, subtract:

\[ 112.36 - 54.76 = 57.6 \]

Finally,

\[ \Delta K.E. = \frac{1}{2} \times 79.5 \times 57.6 \]
\[ \Delta K.E. = 39.75 \times 57.6 \]
\[ \Delta K.E. = 2289.6 \text{ J} \]

Therefore, the work done by the net external force acting on the skier is 2289.6 Joules.

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