COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 5, Problem 80QAP
To determine

The magnitude of the acceleration of the block resting on an inclined plane.

Expert Solution & Answer
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Answer to Problem 80QAP

The magnitude of the acceleration of the block resting on an inclined plane is 3.77 m/s2.

Explanation of Solution

Given info:

Mass of the block placed on the incline

  m1=1.00 kg

Mass of the hanging block

  m2=2.00 kg

Angle made by the plane with horizontal

  θ=30.0°

Coefficient of static friction

  μs=0.500

Coefficient of kinetic friction

  μk=0.400

Formula used:

Free body diagrams are drawn for the two blocks and the acceleration of the block is determined using the force equations for both the blocks.

The free body diagram for the block of mass m2, assuming the positive x axis pointing downwards and the y axis perpendicular to it is shown below.

Since m2 is greater than m1, the hanging block would tend to move down and the block on the incline would slide upwards along the incline.

The weight of the block is w2, which acts along the +x direction and the tension T acts along the −x direction.

  COLLEGE PHYSICS, Chapter 5, Problem 80QAP , additional homework tip  1

The total force acting along the +x direction is given by,

  F2x=w2T=m2ax(1)

Here, ax is the block s acceleration along the downward direction (+x ).

Draw the free body diagram for the block of mass m1. Since the block tends to slide up the plane, the force of friction f acts downwards. The coordinate system under consideration has its +x direction up the incline and the +y direction perpendicular to the incline.

  COLLEGE PHYSICS, Chapter 5, Problem 80QAP , additional homework tip  2

The weight w1 acts downwards and the normal force n acts along the +y direction. The force of friction f acts downwards along the −x direction and the tension T acts parallel to the plane along the +x direction.

Resolve the weight w1 into two components w1x and w1y along the −x and the +y directions.

Both the blocks have the same magnitude of acceleration.

The force equation along the +x direction is given by,

  F1x=Tw1xf=m1ax(2)

The force equation along the +y direction is given by,

  F1y=nw1y(3)

Since the block is in equilibrium in the y direction,

  F1y=nw1y=0

Hence,

  n=w1y.....(4)

The force of friction and the normal force are related as follows:

  f=μn... (5)

The value of the coefficient of friction μ takes the value of μs if the block is at rest and μk if the block is in motion.

Calculation:

First determine, if the system is at rest or in motion.

The system will be at rest if

  T<w1x+fs

If the system is at rest, equation (1) can be written as,

  F2x=w2T=m2ax=0T=w2

Since w2=m2g, where g is the acceleration of free fall

  T=m2g

Substitute the values of the variables in the above equation,

  T=m2g=(2.00 kg)(9.80 m/s2)=19.6 N

Calculate the value of w1x+fs.

The component w1x=w1sinθ, where, w1=m1g.

  w1x=w1sinθ=m1gsinθ=(1.00 kg)(9.80 m/s2)(sin30.0°)=4.90 N

If the system is at rest, assume the maximum force of static friction to act on the block.

Then,

  fs=μsn

From equation (4),

  fs=μsn=μsw1y

The component w1y is given by,

  w1y=w1cosθ, where w1=m1g

Calculate the maximum force of static friction acting on the block.

  fs=μsw1y=μsm1gcosθ=(0.500)(1.00 kg)(9.80 m/s2)(cos30.0°)=4.24 N

Therefore,

  w1x+fs=(4.90 N)+(4.24 N)=9.14 N

Since it is seen that T>w1x+fs, the system accelerates. The force of friction is that of kinetic friction fk.

Add equations (1) w2T=m2ax and (2) Tw1xfk=m1ax

  Tw1xfk=m1axw2w1xfk=(m1+m2)ax

Substitute the expressions for w2,w1x and fk in the expression and write an expression for ax.

  m2gm1gsinθμkm1gcosθ=(m1+m2)ax

  ax=m2gm1gsinθμkm1gcosθ(m1+m2)

Substitute the values of the variables in the expression and determine the magnitude of the acceleration of the block on the incline.

  ax=m2gm1gsinθμkm1gcosθ( m 1+ m 2)=(2.00 kg)(9.80  m/s 2)(1.00 kg)(9.80  m/s 2)(sin30.0°)(0.400)(1.00 kg)(9.80  m/s 2)(cos30.0°)(1.00 kg)+(2.00 kg)=3.77 m/s2

Conclusion:

The magnitude of the acceleration of the block resting on an inclined plane is 3.77 m/s2.

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Chapter 5 Solutions

COLLEGE PHYSICS

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