PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 6, Problem 71P

(a)

To determine

Whether the magnitude of the force is F0 and its direction is perpendicular to r=xi^+yj^ .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The force acting on the particle is (F0r)(yi^xj^) and r is the distance of the particle from the origin.

Formula used:

The vector notation of vector A is A=Axi^+Ayj^ .

Write the expression for magnitude of any vector.

  |A|=(Ax2+Ay2)

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 force.

  |F|=(F0ry)2+(F0rx)2=(F0r)(y2+x2)

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

  |F|=F0

Take the dot product between the force and (xi^+yj^) .

  F.(xi^+yj^)=((F0r)(yi^xj^)).(xi^+yj^)=(F0r)(yxxy)=0

Conclusion:

Thus, the magnitude of the force is F0 and its direction is perpendicular to r=xi^+yj^ .

(b)

To determine

The work done by the force on the particle.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The particle is moving around a circle of radius 5.0m centred at the origin. The force acting on the particle is (F0r)(yi^xj^) .

Formula Used:

Write the expression for the work done by a force on a particle.

  W=θ1θ2F.dr   ........ (1)

Here, W is the work done, F is the force applied, θ1 is the initial angular displacement of the particle and θ2 is the final angular displacement of the particle.

Calculation:

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

  r=(xi^+yj)=rcosθi^+rsinθj^

Here, r is the magnitude of the radial vector and θ is the instantaneous value of the angular position of the paticle.

  dr=(rsinθi^+rcosθj^)dθ

Rearrange the expression for force.

  F=(F0r)(rsinθi^rcosθj^)

Substitute (F0r)(rsinθi^rcosθj^) for F and (rsinθi^+rcosθj^)dθ for r , 0 for θ1 and 2π for θ2 in expression (1).

  W=02π(F0r)(rSinθi^rCosθj^).(rSinθi^+rCosθj^)dθ=F0r02πdθ=2πF0r

Substitute 5.0m for r in the above expression.

  W=(10π)F0

Conclusion:

Thus, the magnitude of the work done by the force on the particle is (10π)F0 .

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