To determine Show that the net work done as W = 1 2 m v 2 2 − 1 2 m v 1 2 by the force in moving object from the distance x 1 to x 2 . Explanation Given information: Mass of the object is m . The equation for the objects velocity v = d v d x . The objects velocities at x 1 and x 2 are v 1 and v 2 . The equation of Newton’s second law of motion is F = m ( d v d t ) . The Chain rule equation is v = d v d t = d v d x d x d t = v d v d x . Calculation: Determine the net work done using the relation. W = ∫ x 1 x 2 F ( x ) d x Substitute m ( d v d t ) for F ( x ) . W = ∫ x 1 x 2 m ( d v d t ) d x Substitute v d v d x for ( d v d t )

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 6.5, Problem 25E

To determine

Show that the net work done as W=12mv22−12mv12 by the force in moving object from the distance x1 to x2.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6.5, Problem 24E

Chapter 6.5, Problem 26E

Students have asked these similar questions

Write an integral that is approximated by the following Riemann sum. Substitute a into the Riemann sum below where a is the last non-zero digit of your banner ID. You do not need to evaluate the integral. 2000 (10 1 ((10-a) +0.001) (0.001)

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter | (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) ☐ 1. For all n > 1, seriesΣ In(n) In(n) converges. 2, 1, arctan(n) the series arctan(n) n³ ☐ 4. For all n > 1, 123 converges. 1 n ln(n) series In(n) diverges. 2n . and the seriesΣconverges, so by the Comparison Test, 2, 3, and the series converges, so by the Comparison Test, the series-3 1 converges. ☐ 6. For all n > 2, In(n) >, and the series Σ converges, so by the Comparison Test, the seriesΣ In(n) converges.

Instructions. "I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."

Chapter 6 Solutions

University Calculus: Early Transcendentals (4th Edition)

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.