EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
bartleby

Concept explainers

Question
Book Icon
Chapter 8, Problem 25AP

(a)

To determine

The constant friction force for the last 20m for the empty test car.

(a)

Expert Solution
Check Mark

Answer to Problem 25AP

The constant friction force for the last 20m for the empty test car is 1.29×104N.

Explanation of Solution

 The mass of the empty car is 250kg , the height of the track is 110m ,the total length of the ground is 250m.

Write the formula to calculate the energy by equation of motion

    Ki+Ui+ΔE=Kf+Uf        (1)

Here, Ki is the initial kinetic energy, Ui is the initial potential energy, ΔE is the energy wasted, Kf is the final kinetic energy and Uf is the final potential energy.

The formula of initial kinetic energy is,

    Ki=12mvi2

Substitute 0 for vi in the above equation to find the value of Ki.

    Ki=12m(02)=0

The formula of kinetic energy at final point

    Kf=12mvf2

Substitute 0 for vf in the above equation to find the value of Kf.

    Kf=12m(02)=0

The formula of the potential energy at start

    Ui=mgh

Substitute yi for h in the above equation to find the value of Ui.

    Ui=mgyi

The formula of the potential energy at final point

    Uf=mgh

Substitute 0 for h in the above equation to find the value of Uf.

    Uf=0

The formula of the energy wasted due to friction

    ΔE=fΔx

Substitute f1Δx1f2Δx2 for fΔx in the above equation to find the value of ΔE.

    ΔE=f1Δx1f2Δx2

Substitute 0 for Ki, mgyi for Ui, f1Δx1f2Δx2 for ΔE, 0 for Uf and 0 for Kf in the equation (1) to find the value of f2 .

    0+mgyif1Δx1f2Δx2=0+0

Conclusion:

Substitute 250kg for m, 9.8m/s2 for g, 110m for yi, 50N for f1, 230m for Δx1 and 20m for Δx2 in the above equation to find the value of f2.

    (250kg)(9.8m/s2)(110m)(50N)(230m)f2(20m)=0f2=12900N=1.29×104N

Therefore, the constant friction force for the last 20m for the empty test car is 1.29×104N.

(b)

To determine

The highest speed reached by the car.

(b)

Expert Solution
Check Mark

Answer to Problem 25AP

The highest speed reached by the car is 45.5m/s.

Explanation of Solution

The formula of kinetic energy at final point

    Kf=12mvf2

The body is at the lowest point so the resistive force is taken upto that point only.

The formula of the energy wasted due to friction

    ΔE=fΔx

Substitute f1Δx1 for fΔx in the above equation to find the value of ΔE.

    ΔE=f1Δx1

Write the formula to calculate the final velocity from equation (1) can be written as,

    mgyif1Δx1=12mvf2        (2)

Conclusion:

Substitute 250kg for m, 9.8m/s2 for g, 110m for yi, 50N for f1 and 230m for Δx1 in equation (2) to find the value of vf.

    (250kg)(9.8m/s2)(110m)(50N)(230m)=12(250kg)vf2125vf2=258000vf=45.5m/s

Therefore, the highest speed reached by the car is 45.5m/s.

(c)

To determine

The new values of friction force and highest speed when the weight is 450kg .

(c)

Expert Solution
Check Mark

Answer to Problem 25AP

The new value of friction force is 3.72×104N and highest speed when the weight is 450kg is 46.1m/s.

Explanation of Solution

 The mass of the empty car is 250kg , the height of the track is 110m ,the total length of the ground is 250m.

The Write the formula to calculate the force of friction form equation (1)

    0+mgyif1Δx1f2Δx2=0+0

Substitute 450kg for m, 9.8m/s2 for g, 110m for yi, 50N for f1, 230m for Δx1 and 20m for Δx2 in the above equation to find the value of f2.

    (450kg)(9.8m/s2)(110m)(50N)(230m)f2(20m)=0f2=37200N=3.72×104N

To calculate the final velocity the equation (2)

    mgyif1Δx1=12mvf2

Substitute 450kg for m, 9.8m/s2 for g, 110m for yi, 50N for f1 and 230m for Δx1 in the above equation to find the value of vf.

    (450kg)(9.8m/s2)(110m)(50N)(230m)1=12(450kg)vf2225vf2=473600vf=46.1m/s

Conclusion:

Therefore, the new value of friction force is 3.72×104N and highest speed when the weight is 450kg is 46.1m/s..

(d)

To determine

The depth of the underground part of the ride.

(d)

Expert Solution
Check Mark

Answer to Problem 25AP

The depth of the underground part of the ride is 45m.

Explanation of Solution

Consider x be the depth of the underground tunnel made beyond height yi

The formula of the potential energy at start

    Ui=mgh

Substitute (yi+x) for h in the above equation to find the value of Ui.

    Ui=mg(yi+x)

The Write the formula to calculate the depth

    mg(yi+x)f1Δx1=12mvf2

Conclusion:

Substitute 450kg for m, 9.85m/s2 for g, 110m for yi, 50N for f1, 55m/s2 for vf and 150m for Δx1 in the above equation to find the value of x.

  (450kg)(9.85m/s2)((110m)+x)(50N)(150m)=12(450kg)(55m/s2)24432.5(110+x)=688125110+x=155.2x=45.245m

x45m

Therefore, the depth of the underground part of the ride is 45m.

(e)

To determine

The depth of the underground provided is feasible or not.

(e)

Expert Solution
Check Mark

Answer to Problem 25AP

No, the depth of the underground provided is not feasible.

Explanation of Solution

No the depth of the underground tunnel provided is not feasible as the total length covered by the car is increased due to which the loss of energy due to friction is increased a lot as compared to the case when the underground path was not there. The non-conventional forces on the car are increased due to the depth provided as the friction in the path of later 20m is very high. So, the depth of the underground provided is not feasible.

Conclusion:

Therefore, the depth of the underground provided is not feasible.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
19:39 · C Chegg 1 69% ✓ The compound beam is fixed at Ę and supported by rollers at A and B. There are pins at C and D. Take F=1700 lb. (Figure 1) Figure 800 lb ||-5- F 600 lb بتا D E C BO 10 ft 5 ft 4 ft-—— 6 ft — 5 ft- Solved Part A The compound beam is fixed at E and... Hình ảnh có thể có bản quyền. Tìm hiểu thêm Problem A-12 % Chia sẻ kip 800 lb Truy cập ) D Lưu of C 600 lb |-sa+ 10ft 5ft 4ft6ft D E 5 ft- Trying Cheaa Những kết quả này có hữu ích không? There are pins at C and D To F-1200 Egue!) Chegg Solved The compound b... Có Không ☑ ||| Chegg 10 וח
No chatgpt pls will upvote
No chatgpt pls will upvote

Chapter 8 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 8 - A crate of mass 10.0 kg is pulled up a rough...Ch. 8 - A 40.0-kg box initially at rest is pushed 5.00 m...Ch. 8 - A smooth circular hoop with a radius of 0.500 m is...Ch. 8 - As shown in Figure P8.10, a green bead of mass 25...Ch. 8 - At time ti, the kinetic energy of a particle is...Ch. 8 - A 1.50-kg object is held 1.20 m above a relaxed...Ch. 8 - Prob. 13PCh. 8 - An 80.0-kg skydiver jumps out of a balloon at an...Ch. 8 - You have spent a long day skiing and are tired....Ch. 8 - The electric motor of a model train accelerates...Ch. 8 - An energy-efficient lightbulb, taking in 28.0 W of...Ch. 8 - An older-model car accelerates from 0 to speed v...Ch. 8 - Make an order-of-magnitude estimate of the power a...Ch. 8 - There is a 5K event coming up in your town. While...Ch. 8 - For saving energy, bicycling and walking are far...Ch. 8 - Energy is conventionally measured in Calories as...Ch. 8 - A block of mass m = 200 g is released from rest at...Ch. 8 - Make an order-of-magnitude estimate of your power...Ch. 8 - Prob. 25APCh. 8 - Review. As shown in Figure P8.26, a light string...Ch. 8 - Consider the blockspringsurface system in part (B)...Ch. 8 - Why is the following situation impossible? A...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - As the driver steps on the gas pedal, a car of...Ch. 8 - As it plows a parking lot, a snowplow pushes an...Ch. 8 - Heedless of danger, a child leaps onto a pile of...Ch. 8 - Review. Why is the following situation impossible?...Ch. 8 - A horizontal spring attached to a wall has a force...Ch. 8 - More than 2 300 years ago, the Greek teacher...Ch. 8 - Review. As a prank, someone has balanced a pumpkin...Ch. 8 - Review. Why is the following situation impossible?...Ch. 8 - An airplane of mass 1.50 104 kg is in level...Ch. 8 - A pendulum, comprising a light string of length L...Ch. 8 - A ball whirls around in a vertical circle at the...Ch. 8 - You are working in the distribution center of a...Ch. 8 - Prob. 43APCh. 8 - Starting from rest, a 64.0-kg person bungee jumps...Ch. 8 - Review. A uniform board of length L is sliding...Ch. 8 - A uniform chain of length 8.00 m initially lies...Ch. 8 - What If? Consider the roller coaster described in...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning