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Modern Physics for Scientists and Engineers

# TEXTBOOK SOLUTIONS FOR Modern Physics for Scientists and Engineers 2nd Edition

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SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 15

(a) The figure below shows the free-body diagram of the block on the inclined. Coordinate axes are shown in red, along and perpendicular to the plane. The sine component of the weight of the block provides the pulling force on the block down the incline, and the cosine component of the weight gives rise to the normal force on the block.

• Step 2 of 15

Position in x -direction:

From the figure, the normal force acting on the block balances the component of the block’s weight, which is given by the expression,

Here, is the weight of the block and is the angle of incline.

Kinetic frictional force acting on the block is equal to the product of the coefficient of kinetic friction and normal force on the block,

Here, is the co-efficient of kinetic friction between the block and the inclined plane.

Substitute for N in this equation,

• Step 3 of 15

The net force on box in x-direction or down the plane is equal to the difference between the sine component of its weight and kinetic frictional force,

Use newton’s law to write the net force equal to the product of mass and acceleration of the block and use equationto substitute the expression for friction here,

This is the expression for the acceleration of the block along the plane or x-axis.

Acceleration of the block is equal to the rate of change in its velocity. Use this to write the above equation as follows:

• Step 4 of 15

Integrating this equation to get the velocity of the block along the plane or x-direction,

Here, is integration constant.

To find the integration constant value, use the condition that the block was at rest initially. So, substitute forin this equation,

Use in the above equation for,

As velocity of the block is equal to the change in its displacement, write this equation as follows:

This gives,
• Step 5 of 15

To find the x position of the block, integrate the above equation,

Here, is integration constant.

As the block was at origin initially, substitute 0 for x and 0 for t,

Use this in the above equation for the position of the block in x-direction,

Therefore, the position of the block in x-direction is.

• Step 6 of 15

Position in y -direction:

The y-axis is perpendicular to the surface of the inclined plane. Recall that the block has no displacement perpendicular to the plane. Hence, its speed and displacement in y-direction are zero.

Therefore, the position of the block y-direction is.

• Step 7 of 15

(b) Free-body diagram for this case of coordinate axes is shown below: weight of the body acts vertically downward (negative y-direction), normal force and friction are also have been resolved into the components along the coordinate axes.

• Step 8 of 15

Position in x -direction:

From the free-body diagram, you can observe that the net force on box in x-direction is equal to the difference between the sine component of normal force and cosine component of the frictional force,

Use newton’s law to write the net force equal to the product of mass and acceleration of the block and use equationandto substitute for N andfor,

This is the expression for the acceleration of the block along the plane or x-axis.

Acceleration of the block is equal to the rate of change in its velocity. Use this to write the above equation as follows:

• Step 9 of 15

Integrating this equation to get the velocity of the block along the plane or x-direction,

Here, is integration constant.

Following the same procedure as down above to find the integration constant, you will see it equal to 0.

Use in the above equation for, which gives,

As velocity of the block is equal to the change in its displacement, write this equation as follows:

This gives,
• Step 10 of 15

To find the x position of the block, integrate the above equation,

Here, is integration constant.

Following the same procedure as earlier, you will find

Use this in the above equation for the position of the block in x-direction,

Therefore, the position of the block in x-direction is.

• Step 11 of 15

Position in y -direction:

From the free-body diagram, you can observe that the net force on box in y-direction is equal to the difference between the weight of the block and sum of the cosine component of normal force and sine component of the frictional force,

Use newton’s law to write the net force equal to the product of mass and acceleration of the block and use equationandto substitute for N andfor,

This is the expression for the acceleration of the block along the plane or y-axis.

Acceleration of the block is equal to the rate of change in its velocity. Use this to write the above equation as follows:

• J Giles
Line 5 (from the bottom) lists the a sub x. However, this is the y-direction portion of the problem.
• Step 12 of 15

Integrating this equation to get the velocity of the block along the plane or x-direction,

Here, is integration constant.

Following the same procedure as down above to find the integration constant, you will see it equal to 0.

Use in the above equation for, which gives,

As velocity of the block is equal to the change in its displacement, write this equation as follows:

This gives,
• Step 13 of 15

To find the y position of the block, integrate the above equation,

Here, is integration constant.

Following the same procedure as earlier, you will find

Use this in the above equation for the position of the block in x-direction,

Therefore, the position of the block in y-direction is.
• Step 14 of 15

In this problem, part (a) and (b) are nothing but the transformation of axes through an angle. So, use the transformation of axes relation as follows:

You need to find the relation between the coordinates (positions) in two cases and show the right hand sides leads to the left hand side, with the same expression as found above.

Use the expressions forandin transformation,

For x -coordinate:

This gives,

This is same as the expression found above in part (a).

• Step 15 of 15

For y -coordinate:

This gives,

This is again same as the expression found above in part (a).

Therefore, both processes give the same final answer.

The main advantages of the Oxy co-ordinate system are that the motion is only along x-direction, friction is along x-direction and the normal force is along y-direction only. Whereas, in co-ordinate system, different directions come in picture and it becomes difficult in simplifying the problem.

Corresponding Textbook

Modern Physics for Scientists and Engineers | 2nd Edition
9780138057152ISBN-13: 013805715XISBN:
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