Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 25AP

(a)

To determine

The constant friction force for the last 20m for the empty test car.

(a)

Expert Solution
Check Mark

Answer to Problem 25AP

The constant friction force for the last 20m for the empty test car is 1.29×104N .

Explanation of Solution

Given info: The mass of the empty car is 250kg , the height of the track is 110m , the total length of the ground is 250m .

The formula to calculate the energy by equation of motion is,

Ki+Ui+ΔE=Kf+Uf (1)

Here,

Ki is the initial kinetic energy.

Ui is the initial potential energy.

ΔE is the energy wasted.

Kf is the final kinetic energy.

Uf is the final potential energy.

The formula of kinetic energy at start is,

Ki=12mvi2

Substitute 0 for vi in the above equation to find the value of Ki .

Ki=12m(02)=0

The formula of kinetic energy at final point is,

Kf=12mvf2

Substitute 0 for vf in the above equation to find the value of Kf .

Kf=12m(02)=0

The formula of the potential energy at start is,

Ui=mgh

Substitute yi for h in the above equation to find the value of Ui .

Ui=mgyi

The formula of the potential energy at final point is,

Uf=mgh

Substitute 0 for h in the above equation to find the value of Uf .

Uf=0

The formula of the energy wasted due to friction is,

ΔE=fΔx

Substitute f1Δx1f2Δx2 for fΔx in the above equation to find the value of ΔE .

ΔE=f1Δx1f2Δx2

Substitute 0 for Ki , mgyi for Ui , f1Δx1f2Δx2 for ΔE , 0 for Uf and 0 for Kf in the equation (1) to find the value of f2 .

0+mgyif1Δx1f2Δx2=0+0

Substitute 250kg for m , 9.8m/s2 for g , 110m for yi , 50N for f1 , 230m for Δx1 and 20m for Δx2 in the above equation to find the value of f2 .

(250kg)(9.8m/s2)(110m)(50N)(230m)f2(20m)=0f2=12900N=1.29×104N

Conclusion:

Therefore, the constant friction force for the last 20m for the empty test car is 1.29×104N .

(b)

To determine

The highest speed reached by the car.

(b)

Expert Solution
Check Mark

Answer to Problem 25AP

The highest speed reached by the car is 45.5m/s .

Explanation of Solution

Given info: The mass of the empty car is 250kg , the height of the track is 110m , the total length of the ground is 250m .

The formula of kinetic energy at final point is,

Kf=12mvf2

The body is at the lowest point so the resistive force is taken upto that point only.

The formula of the energy wasted due to friction is,

ΔE=fΔx

Substitute f1Δx1 for fΔx in the above equation to find the value of ΔE .

ΔE=f1Δx1

The formula to calculate the final velocity from equation (1) can be written as,

mgyif1Δx1=12mvf2 (2)

Substitute 250kg for m , 9.8m/s2 for g , 110m for yi , 50N for f1 and 230m for Δx1 in equation (2) to find the value of vf .

(250kg)(9.8m/s2)(110m)(50N)(230m)=12(250kg)vf2125vf2=258000vf=45.5m/s

Conclusion:

Therefore, the highest speed reached by the car is 45.5m/s .

(c)

To determine

The new values of friction force and highest speed when the weight is 450kg .

(c)

Expert Solution
Check Mark

Answer to Problem 25AP

The new value of friction force is 3.72×104N and highest speed when the weight is 450kg is 46.1m/s .

Explanation of Solution

Given info: The mass of the empty car is 250kg , the height of the track is 110m , the total length of the ground is 250m .

The formula to calculate the force of friction form equation (1) is,

0+mgyif1Δx1f2Δx2=0+0

Substitute 450kg for m , 9.8m/s2 for g , 110m for yi , 50N for f1 , 230m for Δx1 and 20m for Δx2 in the above equation to find the value of f2 .

(450kg)(9.8m/s2)(110m)(50N)(230m)f2(20m)=0f2=37200N=3.72×104N

To calculate the final velocity the equation (2) is,

mgyif1Δx1=12mvf2

Substitute 450kg for m , 9.8m/s2 for g , 110m for yi , 50N for f1 and 230m for Δx1 in the above equation to find the value of vf .

(450kg)(9.8m/s2)(110m)(50N)(230m)1=12(450kg)vf2225vf2=473600vf=46.1m/s

Conclusion:

Therefore, the new value of friction force is 3.72×104N and highest speed when the weight is 450kg is 46.1m/s .

(d)

To determine

The depth of the underground part of the ride.

(d)

Expert Solution
Check Mark

Answer to Problem 25AP

The depth of the underground part of the ride is 45m .

Explanation of Solution

Consider x be the depth of the underground tunnel made beyond height yi

The formula of the potential energy at start is,

Ui=mgh

Substitute (yi+x) for h in the above equation to find the value of Ui .

Ui=mg(yi+x)

The formula to calculate the depth is,

mg(yi+x)f1Δx1=12mvf2

Substitute 450kg for m , 9.85m/s2 for g , 110m for yi , 50N for f1 , 55m/s2 for vf and 150m for Δx1 in the above equation to find the value of x .

(450kg)(9.85m/s2)((110m)+x)(50N)(150m)=12(450kg)(55m/s2)24432.5(110+x)=688125110+x=155.2x=45.245m

x45m

Conclusion:

Therefore, the depth of the underground part of the ride is 45m .

(d)

To determine

The depth of the underground provided is feasible or not.

(d)

Expert Solution
Check Mark

Answer to Problem 25AP

No, the depth of the underground provided is not feasible.

Explanation of Solution

No the depth of the underground tunnel provided is not feasible as the total length covered by the car is increased due to which the loss of energy due to friction is increased a lot as compared to the case when the underground path was not there. The non conventional forces on the car are increased due to the depth provided as the friction in the path of later 20m is very high. So, the depth of the underground provided is not feasible.

Conclusion:

Therefore, the depth of the underground provided is not feasible.

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Physics for Scientists and Engineers with Modern Physics

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