Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

Question
Book Icon
Chapter 8, Problem 8.81CP

(a)

To determine

The minimum speed of the swing with which Jane must begin her swing to just make it to the other side.

(a)

Expert Solution
Check Mark

Answer to Problem 8.81CP

The minimum speed of the swing with which Jane must begin her swing to just make it to the other side is 6.15m/s .

Explanation of Solution

Given info: The mass of Jane is 50.0kg , magnitude of force is 110N , mass of Tarzan is 80.0kg , length of vine  is 40.0m , distance between Tarzan and Jane is D .

The diagram is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 8, Problem 8.81CP

Figure I

From the figure, the width of the river is,

D=Lsinϕ+Lsinθ

Here,

L is the length of the vine.

θ is the angle between vine and the vertical at initial state.

ϕ is the angle between the vine and vertical at the final state.

Rearrange the above formula for ϕ .

D=Lsinϕ+LsinθDLsinθ=Lsinϕsinϕ=DLsinθLϕ=sin1(DLsinθL)

Substitute 50.0m for D , 40.0m for L , 50° for θ in the above formula to find ϕ .

ϕ=sin1(DLsinθL)=sin1(50.0m(40.0m)sin50°50.0m)=sin1(0.484)=28.94°

Thus, the value of ϕ is 28.94° .

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

K.E1=12mv12

Here,

m is the mass of the Jane.

v1 is the initial velocity of Jane.

Thus, the initial kinetic energy of Jane is 12mv12 .

The formula to calculate the final kinetic energy of Jane is,

K.E2=12mv22

Here,

m is the mass of the person.

v2 is the final velocity of the person.

Substitute 0 for v2 in the above formula to find K.E2 .

K.E2=12mv22=12m(0)2=0J

Thus, the final kinetic energy of the Jane is 0J .

The formula to calculate the initial gravitational potential energy is,

P.E1=mg(Lcosθ)

Here,

m is the mass of the Jane.

θ is the angle between the vine and vertical at the initial state.

g is the acceleration due to gravity.

Thus, the initial gravitational potential energy is mgLcosθ .

The formula to calculate the final gravitational potential energy is,

P.E2=mg(Lcosϕ)

Here,

m is the mass of the Jane.

ϕ is the angle between the vine and vertical at final state.

g is the acceleration due to gravity.

Thus the final potential energy of the car is mgLcosϕ .

The formula to calculate the initial work done of the wind due to constant force is,

W=FDcos180°

Here,

F is the constant horizontal force.

D is the width of the river.

Substitute 1 for cos180° in the above formula to find W .

W=FDcos180°=FD(1)=FD

Thus, the initial work done of the wind is FD .

The formula to calculate the law of conservation of energy to the total system is,

K.E1+U1+W=K.E2+U2

Here,

K.E1 is the initial kinetic energy of Jane.

K.E2 is the final kinetic energy of Jane.

U1 is the initial gravitational potential energy.

U2 is the final gravitational potential energy.

W is the initial work done of the wind.

Substitute 12mv12 for K.E1 and 0 for K.E2 , FD for W , mgLcosθ for U1 , mgLcosϕ for U2  in the above formula to find v1 .

K.E1+U1+W=K.E2+U212mv12+mgLcosθ+(FD)=0+(mgLcosϕ)v1=2gLcosθ+2FDmmgLcosϕ

Substitute 110N for F , 9.8m/s2 for g , 50° for θ , 28.94° for ϕ , 50.0kg for m in the above formula to find v1 .

v1=2gLcosθ+2FDmmgLcosϕ=2(9.8m/s2)(40.0m)cos50°+2(110N)(50.0m)(50.0kg)(50.0kg)(9.8m/s2)(40.0m)cos28.94°=6.15m/s

Conclusion:

Therefore, the minimum speed of the swing that must Jane begin her swing to just make it to the other side is 6.15m/s .

(b)

To determine

The minimum speed of the swing at the beginning.

(b)

Expert Solution
Check Mark

Answer to Problem 8.81CP

The minimum speed of the swing at the beginning is 9.87m/s

Explanation of Solution

Given info: The mass of Jane is 50.0kg , magnitude of force is 110N , mass of Tarzan  is 80.0kg , length of vine  is 40.0m , distance between Tarzan and Jane is D .

The formula to calculate the combined mass is,

Mc=mj+mt

Here,

mj is the mass of Jane.

mt is the mass of Tarzan.

Substitute 50.0kg for mj and 80.0kg for mt in the above formula to find Mc .

Mc=mj+mt=(50.0kg)+(80.0kg)=130.0kg

The formula to calculate the initial kinetic energy is,

K.E1=12Mcv12

Here,

Mc is the combined mass.

v1 is the initial velocity.

Thus, the initial kinetic energy is 12Mcv12 .

The formula to calculate the final kinetic energy of Jane is,

K.E2=12Mcv22

Here,

Mc is the combined mass.

v2 is the final velocity .

Substitute 0 for v2 in the above formula to find K.E2 .

K.E2=12Mcv22=12Mc(0)2=0J

Thus, the final kinetic energy is 0J .

The formula to calculate the initial gravitational potential energy is,

P.E1=Mcg(Lcosϕ)

Here,

Mc is the combined mass.

ϕ is the angle between the vine and vertical at the final state.

g is the acceleration due to gravity.

Thus, the initial gravitational potential energy is McgLcosϕ .

The formula to calculate the final gravitational potential energy is,

P.E2=Mcg(Lcosθ)

Here,

Mc is the combined mass.

θ is the angle between the vine and vertical at initial state.

g is the acceleration due to gravity.

Thus the final potential energy of the car is McgLcosθ .

The formula to calculate the initial work done of the wind due to constant force is,

W=FDcos0°

Here,

F is the constant horizontal force.

D is the width of the river.

Substitute 1 for cos0° in the above formula to find W .

W=FDcos0°=FD(1)=FD

Thus, the initial work done of the wind is FD .

The formula to calculate the law of conservation of energy to the total system is,

K.E1+U1+W=K.E2+U2

Here,

K.E1 is the initial kinetic energy of Jane.

K.E2 is the final kinetic energy of Jane.

U1 is the initial gravitational potential energy.

U2 is the final gravitational potential energy.

W is the initial work done of the wind.

Substitute 12Mcv12 for K.E1 and 0 for K.E2 , FD for W , McgLcosϕ for U1 , McgLcosθ for U2 in the above formula to find v1 .

K.E1+U1+W=K.E2+U212Mcv12+McgLcosϕ+(FD)=0+(McgLcosθ)v1=2gLcosϕFDMc2gLcosθ

Substitute 110N for F , 9.8m/s2 for g , 50° for θ , 28.94° for ϕ , 130.0kg for Mc in the above formula to find v1 .

v1=2gLcosϕFDMc2gLcosθ=2(9.8m/s2)(40.0m)cos28.94°(110N)(50.0m)(130.0kg)(50.0kg)(9.8m/s2)(40.0m)cos50°=9.87m/s

Conclusion:

Therefore, the minimum speed of the swing at the beginning is 9.87m/s .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A 70.0 kg circus performer is fired from a cannon that is elevated at an angle of 40 degrees above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 3.00 m from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of 26.8 m between this pint and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.
Tarzan grabs a vine of length L=8.9 m and attempts to swing across the crocodile infested river. The vine breaks just as Tarzan is at lowest point of his swing, a horizontal distance D=60 m and H= 4.5 m above the riverbank. You barely mix it across, landing on the edge of the riverbank. The tension in the vine when it broke was a 1000 N. Find Tarzan's mass.
In order for Jane to return to base camp, she needs to swing across a river of width D that is filled with alligators. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle with the vertical (see below figure). Take D = 50.0 m, F = 113 N, L = 40.0 m, 0 = 47.0°, and her mass to be 50.0 kg. Wind Tarzan L Jane (a) with what minimum speed (in m/s) must Jane begin her swing to just make it to the other side? (If Jane can make it across with zero initial velocity, enter 0.) m/s (b) Shortly after Jane's arrival, Tarzan and Jane decide to swing back across the river (simultaneously). With what minimum speed (in m/s) must they begin their swing? Assume that Tarzan has a mass of 80.0 kg. m/s

Chapter 8 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 8 - In a laboratory model of cars skidding to a stop,...Ch. 8 - What average power is generated by a 70.0-kg...Ch. 8 - A ball of clay falls freely to the hard floor. It...Ch. 8 - A pile driver drives posts into the ground by...Ch. 8 - One person drops a ball from the top of a building...Ch. 8 - A car salesperson claims that a 300-hp engine is a...Ch. 8 - Prob. 8.3CQCh. 8 - Prob. 8.4CQCh. 8 - Prob. 8.5CQCh. 8 - Prob. 8.6CQCh. 8 - In the general conservation of energy equation,...Ch. 8 - Consider the energy transfers and transformations...Ch. 8 - A block is connected to a spring that is suspended...Ch. 8 - In Chapter 7, the work-kinetic energy theorem, W =...Ch. 8 - For each of the following systems and time...Ch. 8 - Prob. 8.2PCh. 8 - A block of mass 0.250 kg is placed on top of a...Ch. 8 - A 20.0-kg cannonball is fired from a cannon with...Ch. 8 - cal energy of the ballEarth sys-tem at the maximum...Ch. 8 - A block of mass m = 5.00 kg is released from point...Ch. 8 - Two objects are connected by a light string...Ch. 8 - Prob. 8.8PCh. 8 - A light, rigid rod is 77.0 cm long. Its top end is...Ch. 8 - At 11:00 a.m, on September 7, 2001, more than one...Ch. 8 - Prob. 8.11PCh. 8 - A sled of mass m is given a kick on a frozen pond....Ch. 8 - A sled of mass m is given a kick on a frozen pond....Ch. 8 - A crate of mass 10.0 kg is pulled up a rough...Ch. 8 - A block of mass m = 2.(K) kg is attached to a...Ch. 8 - A 40.0-kg box initially at rest is pushed 5.00 m...Ch. 8 - A smooth circular hoop with a radius of 0.500 m is...Ch. 8 - At time ti, the kinetic energy of a particle is...Ch. 8 - A boy in a wheelchair (total mass 47.0 kg) has...Ch. 8 - As shown in Figure P8.10, a green bead of mass 25...Ch. 8 - A toy cannon uses a spring to project a 5.30-g...Ch. 8 - The coefficient of friction between the block of...Ch. 8 - A 5.00-kg block is set into motion up an inclined...Ch. 8 - A 1.50-kg object is held 1.20 m above a relaxed...Ch. 8 - A 200-g block is pressed against a spring of force...Ch. 8 - An 80.0-kg skydiver jumps out of a balloon at an...Ch. 8 - Prob. 8.27PCh. 8 - Sewage at a certain pumping station is raised...Ch. 8 - An 820-N Marine in basic training climbs a 12.0-m...Ch. 8 - The electric motor of a model train accelerates...Ch. 8 - When an automobile moves with constant speed down...Ch. 8 - Prob. 8.32PCh. 8 - An energy-efficient lightbulb, taking in 28.0 W of...Ch. 8 - An electric scooter has a battery capable of...Ch. 8 - Make an order-of-magnitude estimate of the power a...Ch. 8 - An older-model car accelerates from 0 to speed v...Ch. 8 - For saving energy, bicycling and walking are far...Ch. 8 - A 650-kg elevator starts from rest. It moves...Ch. 8 - Prob. 8.39PCh. 8 - Energy is conventionally measured in Calories as...Ch. 8 - A loaded ore car has a mass of 950 kg and rolls on...Ch. 8 - Make an order-of-magnitude estimate of your power...Ch. 8 - A small block of mass m = 200 g is released from...Ch. 8 - Prob. 8.44APCh. 8 - Review. A boy starts at rest and slides down a...Ch. 8 - Review. As shown in Figure P8.26, a light string...Ch. 8 - A 4.00-kg particle moves along the x axis. Its...Ch. 8 - Why is the following situation impossible? A...Ch. 8 - A skateboarder with his board can be modeled as a...Ch. 8 - Heedless of danger, a child leaps onto a pile of...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Consider the blockspringsurface system in part (B)...Ch. 8 - As it plows a parking lot, a snowplow pushes an...Ch. 8 - Prob. 8.55APCh. 8 - Prob. 8.56APCh. 8 - As the driver steps on the gas pedal, a car of...Ch. 8 - Review. Why is the following situation impossible?...Ch. 8 - A horizontal spring attached to a wall has a force...Ch. 8 - More than 2 300 years ago, the Greek teacher...Ch. 8 - A child's pogo stick (Fig. P8.61) stores energy in...Ch. 8 - A 1.00-kg object slides to the right on a surface...Ch. 8 - A 10.0-kg block is released from rest at point in...Ch. 8 - Prob. 8.64APCh. 8 - A block of mass 0.500 kg is pushed against a...Ch. 8 - Review. As a prank, someone has balanced a pumpkin...Ch. 8 - Review. The mass of a car is 1 500 kg. The shape...Ch. 8 - A pendulum, comprising a light string of length L...Ch. 8 - A block of mass M rests on a table. It is fastened...Ch. 8 - Review. Why is the following situation impossible?...Ch. 8 - While running, a person transforms about 0.600 J...Ch. 8 - A roller-coaster car shown in Figure P8.72 is...Ch. 8 - A ball whirls around in a vertical circle at the...Ch. 8 - An airplane of mass 1.50 104 kg is in level...Ch. 8 - Prob. 8.75APCh. 8 - In bicycling for aerobic exercise, a woman wants...Ch. 8 - Review. In 1887 in Bridgeport, Connecticut, C. J....Ch. 8 - Prob. 8.78APCh. 8 - Review. A uniform board of length L is sliding...Ch. 8 - Starting from rest, a 64.0-kg person bungee jumps...Ch. 8 - Prob. 8.81CPCh. 8 - Prob. 8.82CPCh. 8 - What If? Consider the roller coaster described in...Ch. 8 - A uniform chain of length 8.00 m initially lies...Ch. 8 - Prob. 8.85CP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning