EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
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Question
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Chapter 8, Problem 13P

(a)

To determine

Whether the child-earth system is isolated or not.

(a)

Expert Solution
Check Mark

Answer to Problem 13P

Therefore, the child-earth system is isolated as the only force that can do work on the child is her weight.

Explanation of Solution

The mass of the child is m, the height of the slide is h and the height from where she launch into the air is h5.

In the child-earth system, the total work on the child is done by the gravitation force (her weight) only as there is no friction and the air resistance is ignored. The work done by the normal force of the slide is zero because her displacement is perpendicular to the normal force.

Hence, the child-earth system is an isolated system.

Conclusion:

Therefore, the child-earth system is isolated as the only force that can do work on the child is her weight.

(b)

To determine

Whether there is a non-conservative force acting within the system or not.

(b)

Expert Solution
Check Mark

Answer to Problem 13P

Therefore, there is no non conservative force acting within the system because there is no frictional force.

Explanation of Solution

Only the weight of the child does work whereas the normal force does no work since the displacement is always perpendicular to the normal force.

The child-earth system is isolated so, there is no loss of energy. There is no non- conservative force acting within the system because frictional force is absent here.

Conclusion:

Therefore, there is no non conservative force acting within the system because there is no frictional force.

(c)

To determine

The total energy of the earth-child system when the child is at the top of the water slide.

(c)

Expert Solution
Check Mark

Answer to Problem 13P

The total energy of the earth-child system when the child is at the top of the water slide is mgh.

Explanation of Solution

Write the formula to calculate the total energy at the top of the water slide

    Et=PEt+KEt        (I)

Here, Et is the total energy of the system at the top of the water slide, PEt is the potential energy of the system at the top of the water slide and KEt is kinetic energy of the system at the top of the water slide.

Write the formula to calculate the potential energy of the system at the top of the water slide

    PEt=mgh

Here, m is the mass of child, g is the acceleration due to gravity and h is the vertical maximum height of the slide.

As the child starts from rest therefore, the kinetic energy at the top is zero.

Substitute 0 for KEt and mgh for PEt in equation (I).

    Et=0+mgh=mgh

Conclusion:

Therefore, the total energy of the earth child system when the child is at the top of the water slide is mgh.

(d)

To determine

The expression for the total energy of the system at the launching point.

(d)

Expert Solution
Check Mark

Answer to Problem 13P

The total energy of the system at the launching point is 12mvi2+mgh5.

Explanation of Solution

The mass of the child is m, the height of the slide is h and the height from where she launch into the air is h5.

Write the formula to calculate the total energy at the launching point of earth child system

    El=PEl+KEl        (II)

Here, El is the total energy of the system at the launching point of the water slide, PEl is the potential energy of the system at the launching point of the water slide and KEl is kinetic energy of the system at the launching point of the water slide.

Write the formula to calculate the potential energy of the system at the launching point of the water slide

    PEl=mgh5

Write the formula to calculate the kinetic energy of the system at the launching point of the water slide

    KEl=12mvi2

Here, vi is the velocity of child at the launching point.

Substitute 12mvi2 for KEl and mgh5 for PEl in equation (II).

    El=12mvi2+mgh5

Conclusion:

Therefore, the total energy of the earth child system at the launching point is 12mvi2+mgh5.

(e)

To determine

The expression for the total energy of the system at the highest point in her projectile motion.

(e)

Expert Solution
Check Mark

Answer to Problem 13P

The total energy of the system at the highest point of the projectile motion is 12mvxi2+mgymax.

Explanation of Solution

Write the formula to calculate the total energy at the highest point in the projectile motion of earth child system

    Eh=PEh+KEh        (III)

Here, Eh is the total energy of the system at the highest point of her projectile motion, PEh is the potential energy of the system at the highest point of her projectile motion and KEh is kinetic energy of the system at the highest point of her projectile motion.

Write the formula to calculate the potential energy of the system at the highest point of her projectile motion

    PEh=mgymax

Here, ymax is the vertical maximum height of her projectile motion.

Write the formula to calculate the kinetic energy of the system at the highest point of her projectile motion

    KEh=12mvh2=12m(vxi2+vyi2)=12m(vxi2+0)=12mvxi2

Here, vh is the velocity of child at the highest point of the projectile motion.

Substitute 12mvxi2 for KEh and mgymax for PEh in equation (III).

    Eh=12mvxi2+mgymax

Conclusion:

Therefore, the total energy of the system at the highest point of the projectile motion is 12mvxi2+mgymax.

(f)

To determine

The speed of child at launching point.

(f)

Expert Solution
Check Mark

Answer to Problem 13P

The speed of child at launching point is vi=22gh5.

Explanation of Solution

As the total energy of the system remains conserved at every point therefore,

    Et=El

Substitute 12mvi2+mgh5 for El and mgh for Et.

    mgh=12mvi2+mgh512vi2=ghgh5vi=22gh5

Conclusion:

Therefore, the speed of child at launching point is vi=22gh5.

(g)

To determine

The maximum airborne height ymax in terms of h and the launch angle θ.

(g)

Expert Solution
Check Mark

Answer to Problem 13P

The maximum airborne height ymax in terms of h and the launch angle θ is h54hsin2θ5.

Explanation of Solution

Write the formula to calculate the maximum height ymax by conservation of energy, the total energy at the highest point is equal to the total energy at the launching point

    Eh=El

Substitute 12mvi2+mgh5 for El and 12mvxi2+mgymax for Eh.

    12mvxi2+mgymax=12mvi2+mgh5        (IV)

The horizontal component of velocity vi

    vxi=vicosθ

Substitute vicosθ for vxi in equation (IV).

    12m(vicosθ)2+mgymax=12mvi2+mgh5ymax=h5vi2sin2θ2g        (V)

Substitute 22gh5 for vi in equation (V).

    ymax=h5+(22gh5)2sin2θ2g=h54hsin2θ5

Conclusion:

Therefore, the speed of child at launching point is h54hsin2θ5.

(h)

To determine

Whether the answers would be same if the waterslide were not frictionless.

(h)

Expert Solution
Check Mark

Answer to Problem 13P

Therefore, the answers would not be same if the waterslide were not frictionless.

Explanation of Solution

Due to the presence of frictional force, the total mechanical energy of the system would not conserve. There would be some frictional losses. So, the kinetic energy of the child at every point after the top of the slide would be less than the kinetic energy when there is no friction.

Thus, less kinetic energy means, her launch speed, maximum height and final speed would also be less.

Conclusion:

Therefore, the answers would not be same if the waterslide were not frictionless.

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Chapter 8 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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