Concept explainers
(a)
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)
Explanation of Solution
Given:
The mass of the particle is 2.0 kg . The force acting on the particle is −(br3)(x ˆi + y ˆj) .
Formula used:
The vector notation of vector →A is →A = Ax ˆi + Ay ˆj .
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| = √(−br3 x)2 + (−br3 y)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 (x ˆi+ y ˆj) .
→F⋅ (x ˆi+ y ˆj) = −((br3)(x ˆi + y ˆj)). (x ˆi+ y ˆj) = − (br3) (x2 + y2)
Substitute r2 for (x2 + y2) in the above expression.
→F⋅ (x ˆi+ y ˆj) = −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)
The work done by the force.
(b)
Explanation of Solution
Given:
The value of b is 3.0 N⋅ m2 .
The particle moves in xy − plane from a point (2.00 m, 0.00 m) to another point (5.00 m , 0.00 m) .
Formula used:
Write the expression of work done by a particle.
W = x2∫x1→F⋅d→s ........ (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 d→s in expression (1).
W = −b x2∫x1dxx2= b [1x]x2x1= b [1x2 − 1x1]
Substitute 5.00 m for x2 , 2.00 m for x1 and 3.0 N⋅m2 for b in the above expression.
W = (3.0 N⋅ m2)(15.00 − 12.00)m-1= − 0.9 N⋅ m
Conclusion:
Thus, the work done by the force is −0.9 N⋅ m .
(c)
The work done by the force on the particle.
(c)
Explanation of Solution
Given:
The particle moving around a circle of radius 7.0 m .
Calculation:
Write the expression for the instantaneous position of the particle on the circumference of the circle.
→r = x ˆi + y ˆj
Write the expression for the work done by the force.
W = (−(br3)(x ˆi + y ˆj)). (x ˆi + y ˆj)= − br3 (x2 + y2)
Substitute r2 for (x2 + y2) in the above expression.
W = −br
Substitute 3.0 N⋅ m2 for b and 7.00 m for r in the above expression.
W = −3 N⋅ m27 m = − 0.428 N⋅ m
Conclusion:
Thus, the work done by the force on the particle moving around a circle is −0.428 N⋅ m .
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