EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804470
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 8, Problem 8.62AP

A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has a speed of vi = 3.00 m/s when it makes contact with a light spring (Fig. P8.62b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Fig. P8.62c). The object is then forced toward the left by the spring (Fig. P8.62d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Fig. P8.62e). Find (a) the distance of compression d, (b) the speed vat the unstretched posi-tion when the object is moving to the left (Fig. P8.624), and (c) the distance D where the abject comes to rest.

Chapter 8, Problem 8.62AP, A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250

Figure P8.62

(a)

Expert Solution
Check Mark
To determine

The distance of the compression.

Answer to Problem 8.62AP

The distance of the compression is 0.378m .

Explanation of Solution

Given info: The mass of the object is 1.0kg , value of coefficient of kinetic friction is 0.250 , speed of the object is 3.00m/s .

The formula to calculate the change in energy is,

ΔE1=μmgd

Here,

m is the mass of the object.

g is the acceleration due to gravity.

μ is the coefficient of kinetic friction.

d is the distance of compression of the spring.

The formula to calculate the initial kinetic energy of the object is,

Ki=12mvi2

Here,

m is the mass of the object.

vi is the initial velocity of the object.

The formula to calculate the final kinetic energy is,

Kf=12mvf2

Here,

m is the mass of the object.

vf is the final velocity of the object.

The formula to calculate initial potential energy is,

Ui=12mxi2

Here,

m is the mass of the block.

xi is the initial compression distance.

Thus the initial potential energy of the block is 12mxi2 .

The formula to calculate the final potential energy is,

Uf=12kxf2

Here,

k is the spring constant.

xf is the final distance of compression.

Thus, the final potential energy of the block is 12kxf2 .

The formula to calculate the initial energy is,

Ei=Ui+Ki

Here,

Ui is the initial potential energy.

Ki is the initial kinetic energy.

Substitute 12mvi2 for Ki and 12kxi2 for Ui in the above formula to find Ei .

Ei=Ui+Kf=12mvi2+12kxi2

Thus, the initial energy is 12mvi2+12kxi2 .

The formula to calculate the final energy is,

Ef=Uf+Kf

Here,

Uf is the final potential energy.

Kf is the final kinetic energy.

Substitute 12mvf2 for Kf and 12kxf2 for Uf in the above formula to find Ef .

Ef=Uf+Kf=12mvf2+12kxf2

Thus, the final energy is 12mvf2+12kxf2 .

The formula to calculate the law of conservation of energy between the second and third diagram is,

EfEi=ΔE1

Here,

Ef is the final energy.

Ei is the initial energy.

Substitute 12mvf2+12kxf2 for Ef , 12mvf2+12kxi2 for Ei , in the above formula to find ΔE .

EfEi=ΔE112mvf2+12kxf212mvi2+mgh=μmgd

Substitute 0 for 12mvf2 , 0 for 12kxi2 and d for xf in the above formula to find d .

12kd212mv12=μmgd

Substitute 1.00kg for m , 9.8m/s2 for g , 0.250  for μ , 50N/m for k  , 3.00m/s for v1 in the above formula to find d .

(0.250)(1.00kg)d=12(50N/m)d2(1.00kg)(3.00m/s)=(25N/m)d2+(2.45N)d4.5Jd=2.45N±(2.45N)24(25.0N)(4.5J)2(25.0N/m)

Further solve the above equation.

d=2.45±21.35N50.0N/m=0.378m

Conclusion:

Therefore, the distance of the compression is 0.378m .

(b)

Expert Solution
Check Mark
To determine

The speed at the unstretched position when the object is moving to the left.

Answer to Problem 8.62AP

The speed at the unstretched position when the object is moving to the left is 2.30m/s .

Explanation of Solution

Given info: The mass of the object is 1.0kg , value of coefficient of kinetic friction is 0.250 , speed of the object is 3.00m/s .

The formula to calculate the change in energy is,

ΔE2=μmg(2d)

Here,

m is the mass of the object.

g is the acceleration due to gravity.

μ is the coefficient of kinetic friction.

d is the distance of compression of the spring.

The formula to calculate the law of conservation of energy between the second and fourth diagram is,

EfEi=ΔE2

Here,

Ef is the final energy.

Ei is the initial energy.

Substitute 12mvf2+12kxf2 for Ef , 12mvi2+12kxi2 for Ei , μmgd(2d) for ΔE2 , in the above formula to find μ .

EfEi=ΔE212mv2+12kx212mv22+mgh=μmgd(2d)

Substitute 0 for 12kxi2 , 0 for 12kxf2 , for in the above formula to find vf .

12mvf212mvi2=μmg(2d)v2=v124(μmg)dm=v124μmgd

Substitute 1.00kg for m , 9.8m/s2 for g , 0.250 for μ , 3.00m/s for vi in the above formula to find v .

vf=vi24μmgd=(3.00m/s)24(0.250)(0.378m)=2.30m/s

Conclusion:

Therefore, the speed at the unstretched position when the object is moving to the left is 2.30m/s .

(c)

Expert Solution
Check Mark
To determine

The distance where the object comes to rest.

Answer to Problem 8.62AP

The distance where the object comes to rest is 1.08m .

Explanation of Solution

Given info: The mass of the object is 1.0kg , value of coefficient of kinetic friction is 0.250 , speed of the object is 3.00m/s .

The formula to calculate the change in energy is,

ΔE3=μmg(D+2d)

Here,

m is the mass of the object.

g is the acceleration due to gravity.

μ is the coefficient of kinetic friction.

d is the distance of compression of the spring.

Thus, the value of change in energy is μmg(D+2d) .

The formula to calculate the law of conservation of energy between the second and fifth diagram is,

EfEi=ΔE3

Here,

Ef is the final energy.

Ei is the initial energy.

Substitute 12mvf2+12kxf2 for Ef , 12mvi2+12kxi2 for Ei , μmgd(D+2d) for ΔE3 , in the above formula to find D .

EfEi=ΔE312mvI2+12kxi212mvf2+mgh=μmgd(D+2d)

Substitute 0 for 12kxi2 , 0 for 12kxf2 , 0 for 12mvf2 in the above formula to find D .

12mvi2=μmg(D+2d)μmg(D+2d)=12mv2D=vi2μg2d

Substitute 1.00kg for m , 9.8m/s2 for g , 0.250 for μ , 3.00m/s for vi in the above formula to find D .

D=vi2μg2d=(3.00m/s)2(0.250)(9.8m/s2)2(0.378)=1.08m

Conclusion:

Therefore, the distance where the object comes to rest is 1.08m .

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Chapter 8 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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