COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 4, Problem 101QAP
To determine

The acceleration of the masses and tension in the strings.

Expert Solution & Answer
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Answer to Problem 101QAP

Tension in String T1=8.40N

Tension in String T2=25.2N

acceleration of the massesa=3.50m/s2

Explanation of Solution

Givendata:

Mass of block A, m1kg=1.00kg

Mass of block B, m2kg=2.00kg

Mass of block B, m4kg=4.00kg

Formula Used:

Newton's second law:

  Fnet=m×a

Calculation:

For theblocks on the ramp, we can use a coordinate system wherethe axes are parallel and perpendicular to the face of theramp and up the ramp is considered positive.

For the 4.00kg block, down will be positive yin order to stay consistent with the ramp's coordinate system.

All three blocks have the sameacceleration because they are joined together.

Using Newton's second law for each of theblocks, we can solve for the acceleration in terms of the known masses.

Once we calculate theacceleration, we can calculate the magnitudes of the tensions.

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 101QAP , additional homework tip  1

Free-body diagram of 1.00kg block:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 101QAP , additional homework tip  2

Free-body diagram of 2.00kgblock:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 101QAP , additional homework tip  3

Free-body diagram of 4.00kg block:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 101QAP , additional homework tip  4

Newton's second law:

  Fext,1-kg,parallel=T1w1kg=T1m1kggsin(30.0)=m1kgaFext,2-kg,parallel=T2T1w2kg=T2T1m2kggsin(30.0)=m2kgaFext,4-kg,y=T2+w4kg=T2+w4kg=T2+m4kgg=m4kga

Solving for the acceleration:

  T2=m4kg(ga)=(m 1kg+m 2kg)a+(m 1kg+m 2kg)gsin(30.0)a=m 4kgg( m 1kg +m 2kg )gsin( 30.0)( m 1kg +m 2kg +m 4kg )=( 4.00kg)( 9.80 m/s2 )( ( 1.00kg )+( 2.00kg ))( 9.80 m/s2 )sin( 30.0)( 1.00kg)+( 2.00kg)+( 4.00kg)=3.50m/s2

Solving for the tensions:

  T1=m1kggsin(30.0)+m1kga=(1.00kg)(9.80 m/s2)sin(30.0)+(1.00kg)(3.50 m/s2)=8.40NT2=m4kg(ga)=(4.00kg)(( 9.80 m/s2 )( 3.50 m/s2 ))=25.2N

Conclusion:

Thus, we havethe acceleration of the masses and tension in the strings as below: -

Tension in String T1=8.40N

Tension in String T2=25.2N

acceleration of the masses a=3.50m/s2

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Chapter 4 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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