Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 77P

(a)

To determine

Speed of the sled and rider.

(a)

Expert Solution
Check Mark

Answer to Problem 77P

Speed is 14.1m/s_.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 7, Problem 77P

Write down the equation of energy conservation.

  Ki+Ugi=Kf+Ugf        (I)

Here Ki is the initial kinetic energy, Ugi is the initial gravitational potential energy, Kf is the final kinetic energy and Ugf is the final gravitational potential energy

Write the equation for initial and final kinetic energy

  Ki=12mvA2Kf=12mvC2        (II)

Here m is the mass, vA is the velocity at point A and vC is the velocity at point C.

Write down the equation for initial and final gravitational potential energy.

  Ugi=mgyAUgf=0        (III)

Here g is the acceleration due to gravity and yA is the height at A.

Substitute (II) and (III) in (I)

  12mvA2+mgyA=12mvC2+0        (IV)

Rearrange (IV) in terms of vC.

    12mvC2=12mvA2+mgyAvC2=vA2+2gyAvC=vA2+2gyA        (V)

Conclusion:

Substitute 2.5m/s for vA, 9.8m/s2 for g and 9.76m for yA in (V)

    vC=(2.5m/s)2+2(9.8m/s2)(9.76m)=14.1m/s

Speed is 14.1m/s_.

(b)

To determine

Magnitude of force on water

(b)

Expert Solution
Check Mark

Answer to Problem 77P

Force is 800N_.

Explanation of Solution

Write the energy conservation equation for a motion incorporated with frictional force

    Ki+Ugifkd=Kf+Ugf        (VI)

Here fk is the frictional force and d is the distance covered.

Substitute (II) and (III) in (VI)

    12mvA2+mgyAfkd=12mvC2+0        (VII)

Final kinetic energy is also zero as the final velocity is zero.

Now rearrange (VII) in terms of fk

    fk=12mvA2+mgyAd        (VIII)

Write the equation for normal force acting

    n=mg        (IX)

Therefore total force on water is,

    F=fk2+n2        (X)

Here F is the magnitude of total force on water.

Conclusion:

Substitute 80kg for m, 2.5m/s for vA, 9.8m/s2 for g, 50m for d and 9.76m for yA in (VIII)

    fk=12(80kg)(2.5m/s)2+(80kg)(9.8m/s2)(9.76m)50m=158N

Substitute 80kg for m and 9.8m/s2 for g in (IX)

    n=(80kg)(9.8m/s2)=784N

Substitute 158N for fk and 784N for n in (X)

    F=(158N)2+(784N)2=800N

Force is 800N_.

(c)

To determine

Force exerted by the chute.

(c)

Expert Solution
Check Mark

Answer to Problem 77P

Force is 771N_.

Explanation of Solution

Use Pythagoras theorem to get the angle of slide

    sinθ=opphypθ=sin1(opphyp)        (XI)

Here θ is the angle of slide, opp is the side opposite to theta, and hyp is the hypotenuse of the right triangle.

Write the equation for force on B

  n=mgcosθ        (XII)

Conclusion:

Substitute 9.76m for opp, 53.4m for hyp in (XI)

    θ=sin1(9.76m54.3m)=10.4°

Substitute 80kg for m, 9.8m/s2 for g and 10.4° for θ in (XIII)

    n=(80kg)(9.8m/s2)cos10.4°=771N

Force is 771N_.

(d)

To determine

Force exerted by chute on point C.

(d)

Expert Solution
Check Mark

Answer to Problem 77P

The force is 1.57×103N_.

Explanation of Solution

Write the equation for net force.

    FC=nCmg        (XIII)

Here nC is the force at C.

Write the equation for centripetal force

    FC=mvC2r        (XIV)

Here r is the radius of the circular path.

Equate (XIII) and (XIV) and write for nC.

    nCmg=mvC2rnC=mg+mvC2r        (XV)

Conclusion:

Substitute 80kg for m, 9.8m/s2 for g, 14.1m/s for vC and 20m for r in (XV)

    nC=(80kg)(9.8m/s2)+(80kg)(14.1m/s)220m=1.57×103N

The force is 1.57×103N_.

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Chapter 7 Solutions

Principles of Physics: A Calculus-Based Text

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