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 74QAP
To determine

(a)

The value of M1 for which the two blocks are in equilibrium (no acceleration)

Expert Solution
Check Mark

Answer to Problem 74QAP

The value of M1=10.0kg for which the two blocks are in equilibrium (no acceleration)

Explanation of Solution

Given data:

  M2=20.0kg

  θ=30.0

Formula Used:

Newton's second law:

  Fnet=m×a

  M(Mass)=T(Force)g(acceleration)

Calculation:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  1

We'll use two different but related coordinate systems for the two blocks. For block M1, positive y will point upward. For block M2, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block M2 does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, both blocks are at rest, which means the acceleration of each block is zero and we can calculate M1 in this case.

.

Free-body diagram of M2 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  2

Newton's second law for M2 :

  Fext,parallel=T+wM2,parallel=M2aparallel=0T=wM2,parallel=M2gsin(30.0)=(20.0kg)(9.80m/s2)sin(30.0)=98.0N

Free-body diagram of M1 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  3

  Fext,y=TwM1=M1ay=0=>wM1=M1g=T=>M1=Tg=98.0N(9.80 m/s 2)=10.0kg

Conclusion:

For two blocks to be in equilibrium the value of M1=10.0kg.

To determine

(b)

The magnitude of acceleration of two blocks if the system can move.

Expert Solution
Check Mark

Answer to Problem 74QAP

The magnitude of acceleration of two blocks 1.96m/s2.

Explanation of Solution

Given data:

  M1=5.0kg

  M2=20.0kg

  θ=30.0

Formula Used:

Newton's second law:

  Fnet=m×a

  M(Mass)=T(Force)g(acceleration)

Calculation:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  4

We'll use two different but related coordinate systems for the two blocks. For block M1, positive y will point upward. For block M2, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block M2 does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, M1=5.00kg, so we can plug this into Newton's second law and calculate the acceleration of the blocks.

The magnitude of the acceleration will be the same for each block because they are joined together with a string.

Free-body diagram of M1 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  5

  Fext,y=TwM1=TM1g=M1ay=M1a=>T=M1(g+a)

Free-body diagram of M2 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  6

Newton's second law for M2 :

  Fext,parallel=T+wM2,parallel=(M1(g+a))+M2gsin(30.0)=M2aparallel=M2a=>a=(M2sin(30.0)M1)gM1+M2=((20.0kg)sin(30.0)(5.0kg))(9.80m/s2)(20.0kg)+(5.0kg)=1.96m/s2

Conclusion:

If the system can move the magnitude of acceleration of two blocks is 1.96m/s2

To determine

(c)

The direction of M2.

Expert Solution
Check Mark

Answer to Problem 74QAP

The direction of M2is down the ramp.

Explanation of Solution

Given data:

  M1=5.0kg

  M2=20.0kg

  θ=30.0

Formula Used:

Newton's second law:

  Fnet=m×a

Calculation:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  7

We'll use two different but related coordinate systems for the two blocks. For block M1, positive y will point upward. For block M2, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block M2 does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, M1=5.00kg, so we can plug this into Newton's second law and calculate the acceleration of the blocks. The magnitude of the acceleration will be the same for each block because they are tethered together with a string.

Free-body diagram of M1 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  8

  Fext,y=TwM1=TM1g=M1ay=M1a=>T=M1(g+a)

Free-body diagram of M2 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  9

Newton's second law for M2 :

  Fext,parallel=T+wM2,parallel=(M1(g+a))+M2gsin(30.0)=M2aparallel=M2a=>a=(M2sin(30.0)M1)gM1+M2=((20.0kg)sin(30.0)(5.0kg))(9.80m/s2)(20.0kg)+(5.0kg)=1.96m/s2

Since a=1.96m/s2 is positive, M2 moves down the ramp.

Conclusion:

As we have value of a as positive, so, M2 moves down the ramp.

To determine

(d)

The distance to which block M2 move in 2.00s.

Expert Solution
Check Mark

Answer to Problem 74QAP

The distance to which block M2move in 2.00sis 3.92m

Explanation of Solution

Given data:

  M1=5.0kg

  M2=20.0kg

  θ=30.0

Formula Used:

Newton's second law:

  Fnet=m×a

And x=ut+12at2

Calculation:

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  10

We'll use two different but related coordinate systems for the two blocks. For block M1, positive y will point upward. For block M2, the axes will be parallel and perpendicular to the inclined plane, where down the ramp and out of the ramp are positive.

We can draw free-body diagrams and apply Newton's second law in component form for both blocks.

The tension acting on each block will be identical in magnitude.

Since block M2 does not leave the plane, the acceleration in the perpendicular direction is always zero.

Since, M1=5.00kg, so we can plug this into Newton's second law and calculate the acceleration of the blocks. The magnitude of the acceleration will be the same for each block because they are tethered together with a string.

If the acceleration is constant, we can use the constant acceleration equations to calculate how far M2 travels starting from rest.

Free-body diagram of M1 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  11

  Fext,y=TwM1=TM1g=M1ay=M1a=>T=M1(g+a)

Free-body diagram of M2 :

  COLLEGE PHYSICS LL W/ 6 MONTH ACCESS, Chapter 4, Problem 74QAP , additional homework tip  12

Newton's second law for M2 :

  Fext,parallel=T+wM2,parallel=(M1(g+a))+M2gsin(30.0)=M2aparallel=M2a=>a=(M2sin(30.0)M1)gM1+M2=((20.0kg)sin(30.0)(5.0kg))(9.80m/s2)(20.0kg)+(5.0kg)=1.96m/s2

By using the constant acceleration equations to calculate how far M2 travels starting from rest.

  x=x0+v0,parallelt+12aparallelt2=0+0+12(1.96m/s2)(2.00s)2=3.92m

Conclusion:

Thus in 2.00s block M2 moves 3.92m of distance.

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Chapter 4 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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