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 64P

(a)

To determine

The work done by the force acting on the particle and the velocity of the particle.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The force acting on the particle is (2.0N/m2)x2i^ . The particle moves along a straight line from point (2.0m,2.0m) topoint (2.0m,7.0m) . The particle starts moving from rest.

Formula used:

Write the expression of work done by a particle.

  W=x1x2F.ds   ........ (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.

Write the expression for the work-energy theorem.

  W=KfKi   ........ (2)

Here, W is the work done, Kf is the final kinetic energy of the particle and Ki is the initial kinetic energy of the particle.

Calculation:

Write the expression for the displacement of the particle.

  ds=(x2x1)i^+(y2y1)j^

Here, ds is the magnitude of displacement, (x1,y1) are the initial position coordinates of the particle and (x2,y2) are the final position coordinates of the particle.

Substitute 0m for (x2x1) and 5m for (y2y1) in the above expression.

  ds=5j^

Substitute (2.0N/m2)x2i^ for F , 5j^ for dx , 2 for x1 and 7 for x2 inexpression (1).

  W=27((2.0N/m2)x2i^).(5j^)=0

Write the expression for the kinetic energy of the particle.

  K=12mv2

Here, k is the kinetic energy, m is the mass of the particle and v is the velocity of the particle.

Write the expression of the final kinetic energy of the particle.

  Kf=12mvf2   ........ (3)

Here, vf is the final velocity of the particle and Kf is the final kinetic energy.

Substitute 12mvf2 for Kf , 0 for Ki and 0 for W in expression (2).

  0=12mvf20

Rearrange the above expression.

  vf=0

Conclusion:

Thus, the work done is 0 and the final speed of the particle is 0.

(b)

To determine

The work done by the force acting on the particle and the velocity of the particle.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The particle moves along a straight line from point (2.0m,2.0m) to point (5.0m,6.0m) .

Calculation:

Write the expression for the displacement of the particle.

  ds=(dx)i^+(dy)j^

Here, dx is the magnitude of displacement, dx is the displacement of the particle along x axis and dy is the displacement of the particle along y axis.

Substitute (2.0N/m2)x2i^ for F , (dx1i^+dy1j^) for dx , 2 for x1 and 5 for x2 inexpression (1).

  W=25((2.0N/m2)x2i^).(dxi^+dyj^)=(2.0N/m2)25x2dx=(2.0N/m2)[x33]25=78J

Substitute 12mvf2 for Kf , 0 for Ki and 78J for W in expression (2).

  78J=12mvf20

Rearrange the above expression.

  vf=2(78J)m=156Jm

Here, m is the mass of the particle.

Conclusion:

Thus, the work done is 78J and the final speed of the particle is 156Jm .

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