COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5, Problem 81QAP
To determine

(a)

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100.

Expert Solution
Check Mark

Answer to Problem 81QAP

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100 is 8.12 s.

Explanation of Solution

Given:

The length of the ski slope

  Δx=250 m

The angle made by the slope to the horizontal

  θ=37.0°

Initial speed of the ski

  v0=10.0 m/s

Coefficient of kinetic friction between the ski and snow

  μk=0.100

Formula used:

A free body diagram of the ski is drawn to analyze its motion.

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 5, Problem 81QAP

Assume a coordinate system, with the +x directed downwards along the incline and +y directed upwards, perpendicular to the incline. The weight w acts vertically downwards, the normal force n acts perpendicular to the incline along the +y direction. The force of kinetic friction fk acts upwards along the incline along the −x direction.

Resolve the weight w along the +x and the −y directions. Use the expression w=mg, where g is the acceleration of free fall and write expressions for the components.

  wx=wsinθ=mgsinθwy=wcosθ=mgcosθ......(1)

The ski is in equilibrium along the y direction.

Therefore,

  Fy=nwy=0

Therefore, using equation (1),

  n=wy=mgcosθ......(2)

The force of kinetic friction and the normal force are related according to the following equation:

  fk=μkn

From equation (2)

  fk=μkn=μkmgcosθ......(3)

Write the force equation along the +x direction.

  Fx=wxfk=max

Use the values of wx and fk from equations (2) and (3) in the expression,

  wxfk=maxmgsinθμkmgcosθ=max

Simplify and write an expression for ax.

  ax=g(sinθμkcosθ)......(4)

Use the following equation of motion to obtain the value of the time t.

  Δx=v0t+12axt2......(5)

Calculation:

Substitute the values of the variables in equation (4) and calculate the value of the acceleration.

  ax=g(sinθμkcosθ)=(9.80 m/s2)[(sin37.0°)(0.100)(cos37.0°)]=5.12 m/s2

Use the calculated value of ax and the values of v0 and Δx in equation (5).

  (250 m)=(10.0 m/s)t+12(5.12 m/s2)t2

Solve the quadratic equation.

  (2.56 m/s2)t2+(10.0 m/s)t+(250 m)=0

  t=(10.0 m/s)± ( 10.0 m/s ) 24( 2.56  m/s 2 )( 250 m)2(2.56  m/s 2)=(10.0 m/s)±(51.58 m/s)(5.12  m/s 2)

Taking the positive root,

  t=8.12 s

Conclusion:

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.100 is 8.12 s.

To determine

(b)

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150.

Expert Solution
Check Mark

Answer to Problem 81QAP

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150 is 8.39 s.

Explanation of Solution

Given:

The length of the ski slope

  Δx=250 m

The angle made by the slope to the horizontal

  θ=37.0°

Initial speed of the ski

  v0=10.0 m/s

Coefficient of kinetic friction between the ski and snow

  μk=0.150

Formula used:

The acceleration of the ski down the slope is given by

  ax=g(sinθμkcosθ)

The time taken to reach the bottom of the slope is calculated using the expression,

  Δx=v0t+12axt2

Calculation:

Substitute the values of the variables in equation for acceleration and calculate the value of the acceleration.

  ax=g(sinθμkcosθ)=(9.80 m/s2)[(sin37.0°)(0.150)(cos37.0°)]=4.72 m/s2

Use the calculated value of ax and the values of v0 and Δx in equation of motion.

  (250 m)=(10.0 m/s)t+12(4.72 m/s2)t2

Solve the quadratic equation.

  (2.36 m/s2)t2+(10.0 m/s)t+(250 m)=0

  t=(10.0 m/s)± ( 10.0 m/s ) 24( 2.36  m/s 2 )( 250 m)2(2.36  m/s 2)=(10.0 m/s)±(49.6 m/s)(4.72  m/s 2)

Taking the positive root,

  t=8.39 s

Conclusion:

The time taken by the runaway ski to slide down a 250-m long slope when the coefficient of kinetic friction between the ski and the snow is 0.150 is 8.39 s.

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
a cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?
Calculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were:  222.22 800.00 61.11 641.67 0.00 588.89 11.11 588.89 8.33 588.89 11.11 588.89 5.56 586.11 2.78 583.33   Give in the answer window the calculated repeated experiment variance in m/s2.
No chatgpt pls will upvote

Chapter 5 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

Ch. 5 - Prob. 11QAPCh. 5 - Prob. 12QAPCh. 5 - Prob. 13QAPCh. 5 - Prob. 14QAPCh. 5 - Prob. 15QAPCh. 5 - Prob. 16QAPCh. 5 - Prob. 17QAPCh. 5 - Prob. 18QAPCh. 5 - Prob. 19QAPCh. 5 - Prob. 20QAPCh. 5 - Prob. 21QAPCh. 5 - Prob. 22QAPCh. 5 - Prob. 23QAPCh. 5 - Prob. 24QAPCh. 5 - Prob. 25QAPCh. 5 - Prob. 26QAPCh. 5 - Prob. 27QAPCh. 5 - Prob. 28QAPCh. 5 - Prob. 29QAPCh. 5 - Prob. 30QAPCh. 5 - Prob. 31QAPCh. 5 - Prob. 32QAPCh. 5 - Prob. 33QAPCh. 5 - Prob. 34QAPCh. 5 - Prob. 35QAPCh. 5 - Prob. 36QAPCh. 5 - Prob. 37QAPCh. 5 - Prob. 38QAPCh. 5 - Prob. 39QAPCh. 5 - Prob. 40QAPCh. 5 - Prob. 41QAPCh. 5 - Prob. 42QAPCh. 5 - Prob. 43QAPCh. 5 - Prob. 44QAPCh. 5 - Prob. 45QAPCh. 5 - Prob. 46QAPCh. 5 - Prob. 47QAPCh. 5 - Prob. 48QAPCh. 5 - Prob. 49QAPCh. 5 - Prob. 50QAPCh. 5 - Prob. 51QAPCh. 5 - Prob. 52QAPCh. 5 - Prob. 53QAPCh. 5 - Prob. 54QAPCh. 5 - Prob. 55QAPCh. 5 - Prob. 56QAPCh. 5 - Prob. 57QAPCh. 5 - Prob. 58QAPCh. 5 - Prob. 59QAPCh. 5 - Prob. 60QAPCh. 5 - Prob. 61QAPCh. 5 - Prob. 62QAPCh. 5 - Prob. 63QAPCh. 5 - Prob. 64QAPCh. 5 - Prob. 65QAPCh. 5 - Prob. 66QAPCh. 5 - Prob. 67QAPCh. 5 - Prob. 68QAPCh. 5 - Prob. 69QAPCh. 5 - Prob. 70QAPCh. 5 - Prob. 71QAPCh. 5 - Prob. 72QAPCh. 5 - Prob. 73QAPCh. 5 - Prob. 74QAPCh. 5 - Prob. 75QAPCh. 5 - Prob. 76QAPCh. 5 - Prob. 77QAPCh. 5 - Prob. 78QAPCh. 5 - Prob. 79QAPCh. 5 - Prob. 80QAPCh. 5 - Prob. 81QAPCh. 5 - Prob. 82QAPCh. 5 - Prob. 83QAPCh. 5 - Prob. 84QAPCh. 5 - Prob. 85QAPCh. 5 - Prob. 86QAPCh. 5 - Prob. 87QAPCh. 5 - Prob. 88QAPCh. 5 - Prob. 89QAPCh. 5 - Prob. 90QAP
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
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY