FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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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.

  FlipIt for College Physics (Algebra Version - Six Months 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.

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Chapter 5 Solutions

FlipIt for College Physics (Algebra Version - Six Months 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
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