×
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 11 - Problem 10p
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 11 - Problem 10p

×

# At what displacement from equilibrium is the speed of a ISBN: 9780130606204 3

## Solution for problem 10P Chapter 11

Physics: Principles with Applications | 6th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants Physics: Principles with Applications | 6th Edition

4 5 1 297 Reviews
22
2
Problem 10P

At what displacement from equilibrium is the speed of a SHO half the maximum value?

Step-by-Step Solution:

SolutionStep 1Listing the data Amount of gas, m = 1 gal = 3.2 kg Speed of bicycle, v = 15 km/h Energy in 1 g gasoline, E = 44 kJ 1 kJ = 1000 JStep 2Converting weight to energy equivalent 1 gal = 3.2 kg 1 kg = 1000 g Then, 3.2 kg = 3.2 × 1000 g = 3200 g Energy in 1 g gasoline, E (1 g) = 44 kJ Energy in X g gasoline, E (X g) = X × 44 kJ Here, X = 3200 Therefore, E (3200 g) = 3200 × 44 kJ = E (1 gal) That is, E (1 gal) = E (3200 g) = 140800 kJ = 1.408 × 10 kJ5 5 8 Or E (1 gal) = E (3200 g) = 1.408 × 10 ×1000 J = 1.408 × 10 J.Step 3Calculating the time for travel from power-energy equation\n The power of a system is, P = E / t Where, E - Energy produced t - time Rearranging the equation, t = E / P Provided, P = 480 W (to pedal a bicycle at 15 km/h speed) t = (1.408 × 10 J) / 480 W = 2.93333 × 10 J / W The unit watts, W = J/s Therefore, t = 2.93333 × 10 J / J/s = 293333 seconds t = 293333...

Step 2 of 3

Step 3 of 3

##### ISBN: 9780130606204

Unlock Textbook Solution