EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 6, Problem 73P

(a)

To determine

The x -component of the force as a function of x.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the block is m . The vertical distance of the top of the block and the ceiling is y0 and the horizontal displacement is x . Spring constant is k .

Formula used:

Write the expression for the force exerted by a spring to a particle.

  F=kx

Here, F is the force applied, k is spring constant and x is the displacement of the particle at any instant of time.

Calculation:

Write the expression for the force exerted on the block by the spring.

  F=kz   ........ (1)

Here, z is the distance by which the spring is stretched.

Write the expression for z .

  z=ly0   ........ (2)

Here, l is the length of the spring.

Write the expression for l .

  l=(x2+y02)

Substitute (x2+y02) for l in expression (2).

  z=(x2+y02)y0

Substitute ((x2+y02)y0) for z in expression (1).

  F=k((x2+y02)y0)

Write the expression for the x component of the force.

  Fx=Fcosθ   ........ (3)

Substitute x(x2+y02) for cosθ in expression (3).

  Fx=Fx(x2+y02)   ........ (4)

Substitute k((x2+y02)y0) for F in expression (4).

  Fx=Fx(x2+y02)=k((x2+y02)y0)x(x2+y02)=kx(1y0(x2+y02))   ........(5)

Conclusion:

Thus, the x-component of the force is kx(1y0(x2+y02)).

(b)

To determine

Whether the x component of the force is proportional to x3 for sufficiently small values of |x| .

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Rearrange the expression (5).

  Fx=kx(11(1+(xy0)2))=kx(1(1+(xy0)2)1/2)

Expand the above expression in Taylor’s series and approximate up to second-order term.

  Fxkx(1112(xy0)2)=kx(12x2y02)=kx32y02

Conclusion:

Thus, the x component of the force is proportional to x3 for sufficiently small values of |x| .

(c)

To determine

The speed of the block.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The block is released from rest at x=x0 , where |x|<<y0 .

Formula Used:

Write the expression for the relation between final velocity, initial velocity, acceleration and the displacement of a particle.

  v2=u2+2aS

Here, v is the final velocity, u is the initial velocity, a is the acceleration and S is the displacement of the particle.

Calculation:

Write the expression for the acceleration of the particle.

  a=Fxm

Substitute kx32y02 for Fx in the above expression.

  a=kx32my02

Write the expression for the velocity of the particle when it crosses x=0 starting from x=x0 at rest.

  v=2ax0

Substitute kx032my02 for a in the above expression.

  v=kx04my02

Conclusion:

Thus, the speed of the block is kx04my02 .

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