Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 6, Problem 72P

(a)

To determine

To Show: The magnitude of the force is inversely proportional to r2 and the direction of the force is opposite to the radius vector.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the particle is 2.0kg . The force acting on the particle is (br3)(xˆi+yˆj) .

Formula used:

The vector notation of vector A is A=Axˆi+Ayˆj .

Write the expression for magnitude of any vector.

  |A|=(A2x+A2y)

Here, Ax is the x component of the vector and Ay is the y component of the vector.

Calculation:

Write the expression for the magnitude of the given force.

  |F|=(br3x)2+(br3y)2=(br3)(x2+y2)

Substitute r for (x2+y2) in the above expression.

  |F|=br2

From the above expression it can be concluded that force is inversely proportional to square of radius.

  |F|1r2

Take the dot product between the force and (xˆi+yˆj) .

  F(xˆi+yˆj)=((br3)(xˆi+yˆj)).(xˆi+yˆj)=(br3)(x2+y2)

Substitute r2 for (x2+y2) in the above expression.

  F(xˆi+yˆj)=br   ........ (1)

The direction is in the opposite to r .

Conclusion:

Thus, the magnitude of the force is inversely proportional to r2 and the direction of the force is opposite to the radius vector.

(b)

To determine

The work done by the force.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The value of b is 3.0Nm2 .

The particle moves in xy plane from a point (2.00m, 0.00m) to another point (5.00m ,0.00m) .

Formula used:

Write the expression of work done by a particle.

  W=x2x1Fds   ........ (1)

Here, W is the work done, F is the force acting on the particle, x1 is the initial position of the particle and x2 is the final position of the particle.

Calculation:

The particle is moving along x axis.

Substitute bx2 for the magnitude of F and dx for the magnitude of ds in expression (1).

  W=bx2x1dxx2=b[1x]x2x1=b[1x21x1]

Substitute 5.00m for x2 , 2.00m for x1 and 3.0Nm2 for b in the above expression.

  W=(3.0Nm2)(15.0012.00)m-1=0.9Nm

Conclusion:

Thus, the work done by the force is 0.9Nm .

(c)

To determine

The work done by the force on the particle.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The particle moving around a circle of radius 7.0m .

Calculation:

Write the expression for the instantaneous position of the particle on the circumference of the circle.

  r=xˆi+yˆj

Write the expression for the work done by the force.

  W=((br3)(xˆi+yˆj)).(xˆi+yˆj)=br3(x2+y2)

Substitute r2 for (x2+y2) in the above expression.

  W=br

Substitute 3.0Nm2 for b and 7.00m for r in the above expression.

  W=3Nm27m=0.428Nm

Conclusion:

Thus, the work done by the force on the particle moving around a circle is 0.428Nm .

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