Chapter 36: Q. 51 (page 1061)

The quantity dE/dv, the rate of increase of energy with speed, is the amount of additional energy a moving object needs per $1 m / s$ increase in speed.
a. $A 25, 000 k g$ kg truck is traveling at $30 m / s$ . How much additional energy is needed to increase its speed by $1 m / s$ ?
b. $A 25, 000 k g$ rocket is traveling atrole="math" localid="1649677033976" $0.90 c$ . How much additional energy is needed to increase its speed by $1 m / s$ ?

Short Answer

Expert verified

a. $750 K J$ energy is needed to increase its speed by $1 m / s$ .

b. $8.2 \times 10^{13} J$ energy is needed to increase its speed by $1 m / s$

Step by step solution

Part (a) step 1: Given Information

We need to find the additional energy needed to increase its speed by $1 m / s$ in which his truck is traveling at $30 m / s$ .

Part (a) step 2: Simplify

Consider

$m = 25, 000 k g$

Here the speed is not relativistic $(v = 30 m / s)$

role="math" localid="1649678026406" $E = \frac{1}{2} m v^{2} \frac{d E}{d v} = m v \frac{d E}{d v} = (25, 000 k g) (30 m / s) \frac{d E}{d v} = 750 K J$ (Substitute values in the equation )

Part (b) step 1: Given Information

We need to find the additional energy needed to increase its speed by $1 m / s$ in witch rocket is traveling at $0.90 c$ .

Part (b) step 2: Simplify

Here the speed is definitely( $v = 0.90 c m / s$ ):

$\frac{d E}{d v} = \frac{d}{d v} (m c^{2} + K) \frac{d E}{d v} = \frac{d}{d v} (m c^{2} + γ_{p} m c^{2} - m c^{2}) \frac{d E}{d v} = \frac{d}{d v} (γ_{p} m c^{2}) \frac{d E}{d v} = m c^{2} \frac{d}{d v} (1 - \frac{v^{2}}{c^{2}}) \frac{d E}{d v} = m c^{2} \frac{d}{d v} {(1 - \frac{v^{2}}{c^{2}})}^{- \frac{1}{2}} \frac{d E}{d v} = m c^{2} (- \frac{1}{2}) {(1 - \frac{v^{2}}{c^{2}})}^{- \frac{3}{2}} (\frac{- 2 v}{c^{2}}) \frac{d E}{d v} = m v {(1 - \frac{v^{2}}{c^{2}})}^{^{- \frac{3}{2}}} \frac{d E}{d v} = (25000 k g) (0.90) (3.0 \times 10^{8} m / s) {(1 - 0 . 9^{2})}^{- \frac{3}{2}} \frac{d E}{d v} = 8.2 \times 10^{13} J$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Part (a) step 1: Given Information

Part (a) step 2: Simplify

Part (b) step 1: Given Information

Part (b) step 2: Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Nuclear Physics

Electromagnetism

Magnetism and Electromagnetic Induction

Circular Motion and Gravitation

Rotational Dynamics

Electricity and Magnetism

Study anywhere. Anytime. Across all devices.