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 4, Problem 93QAP
To determine

(a)

The tension in the rope.

Expert Solution
Check Mark

Answer to Problem 93QAP

The tension in the rope is 644N

Explanation of Solution

Givendata:

Distance, Δy=2.00m

Time, t=10.0s

Mass of Sue, mSue=66.0kg

Formula Used:

Newton's second law:

  Fnet=m×a

  v=ut+12at2

Calculation:

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 4, Problem 93QAP , additional homework tip  1

We'll use two different but related coordinate systems for the two people.

For Sue, positive y will point upward.

For Paul, the axes will be parallel and perpendicular to the inclined plane, where up the ramp and out of the ramp are positive.

Tension from the rope pulling up and gravity pulling down are the only forces acting on

Sue.

Assuming her acceleration is constant, we can use the constant acceleration equations

and Newton's second law to calculate the magnitude of the tension.

Sue's acceleration:

  y=y0+v0yt+12ayt2=y0+0+12ayt2=>ay=2(Δy)t2=2(2.00m) (10.0s)2=0.0400m/s2

Free-body diagram of Sue:

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 4, Problem 93QAP , additional homework tip  2

Newton's second law for Sue:

  Fext,y=TwSue=TmSueg=mSueayT=mSue(g+ay)=(66.0kg)(( 9.80 m/s2 )+( 0.0400 m/s2 ))=644N

Conclusion:

Thus, from theNewton's second law for Sue we have the tension in the rope joining them as 644N

To determine

(b)

Mass of Paul

Expert Solution
Check Mark

Answer to Problem 93QAP

Mass of Paul is 92.4kg

Explanation of Solution

Given data:

Distance, Δy=2.00m

Time, t=10.0s

Mass of Sue, mSue=66.0kg

Formula Used:

Newton's second law:

  Fnet=m×a

  v=ut+12at2

Calculation:

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 4, Problem 93QAP , additional homework tip  3

We'll use two different but relatedcoordinate systems for the two people.

For Sue,positive y will point upward.

For Paul, the axes willbe parallel and perpendicular to the inclined plane,where up the ramp and out of the ramp are positive.

Tension from the rope pulling up and gravity pulling down are the only forces acting on

Sue.

Assuming her acceleration is constant, we can use the constant acceleration equations

and Newton's second law to calculate the magnitude of the tension.

Since Paul and Sue aretethered to one another, the magnitudes of their accelerations are equal.

The tension in therope and gravity are the only forces acting on Paul that have components that are parallel tothe face of the glacier. We can then solve the parallel component of Newton's second law forPaul's mass.

Sue's acceleration:

  y=y0+v0yt+12ayt2=y0+0+12ayt2=>ay=2(Δy)t2=2(2.00m) (10.0s)2=0.0400m/s2

Free-body diagram of Sue:

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 4, Problem 93QAP , additional homework tip  4

Newton's second law for Sue:

  Fext,y=TwSue=TmSueg=mSueayT=mSue(g+ay)=(66.0kg)(( 9.80 m/s2 )+( 0.0400 m/s2 ))=644N

Free-body diagram of Paul:

  FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 4, Problem 93QAP , additional homework tip  5

Newton's second law for Paul:

  Fext,parallel=TwPaul=TmPaulgsin(45.0)=mPaulaparallel=>mPaul=Ta parallel+gsin( 45.0)=644N( 0.0400 m/s2 )+( 9.80 m/s2 )sin( 45.0)=92.4kg

Conclusion:

Thus, by Newton's second law for Paul mass of Paul is 92.4kg

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Chapter 4 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

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