EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100460300
Author: SERWAY
Publisher: YUZU
bartleby

Concept explainers

Question
Book Icon
Chapter 8, Problem 8.82CP

(a)

To determine

The expression for H as a function of F .

(a)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The expression for H as a function of F is 1.61+8.643N2/F2 .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

The diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 8, Problem 8.82CP , additional homework tip  1

Figure I

The formula to calculate the horizontal displacement is,

y=L2(LH)2

Here,

L is the length of the string.

H is the height of the ball from the ground.

The formula to calculate the work done by the wind force is,

W=Fy

Here,

F is the mass of the rider.

y is the horizontal displacement.

Substitute L2(LH)2 for y in the above formula to find W .

W=Fy=F(L2(LH)2)

The formula to calculate the gravitational potential energy of the ball is,

U=mgH

Here,

m is the mass of the ball.

g is the acceleration due to gravity.

H is the height of the ball from the ground.

From the law of conservation of energy,

W=U

Here,

W is the work done by the wind force.

U is the gravitational potential energy of the ball.

Substitute F(L2(LH)2) for W and mgH for U in the above formula to find H .

F(L2(LH)2)=mgH

Square the above expression on both sides to find H .

F2(L2(LH)2)=m2g2H2F2(L2L2H2+2LH)=m2g2H2F2(2LH)=(m2g2+F2)H2H=F2(2L)(m2g2+F2)

Substitute 300g for m , 80.0cm for L and 9.8m/s2 for g in the above formula to find H .

H=F2(2(80.0cm×102m1cm))((300g×103kg1g)2(9.8m/s2)2+F2)=1.6F2(8.643N+F2)=1.61+8.643N2/F2

Conclusion:

Therefore, the expression for H as a function of F is 1.61+8.643N2/F2 .

(b)

To determine

The value of H for F=1.00N .

(b)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of H for F=1.00N is 0.166m .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for H in terms of F is,

H=F2(2L)(m2g2+F2)

Substitute 300g for m 9.8m/s2 for g , 80.0cm for L and 1.00N for F in the above formula to find H .

H=(1.00N)2(2(80.0cm×102m1cm))((300g×103kg1g)2(9.8m/s2)2+(1.00N)2)=1.6Nm9.6436N=0.1659m0.166m

Conclusion:

Therefore, the value of H for F=1.00N is 0.166m .

(c)

To determine

The value of H for F=10.0N .

(c)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of H for F=10.0N is 1.4727m .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for H in terms of F is,

H=F2(2L)(m2g2+F2)

Substitute 300g for m 9.8m/s2 for g , 80.0cm for L and 10.0N for F in the above formula to find H .

H=(10.0N)2(2(80.0cm×102m1cm))((300g×103kg1g)2(9.8m/s2)2+(10.0N)2)=160Nm108.6436N=1.4727m

Conclusion:

Therefore, the value of H for F=10.0N is 1.4727m .

(d)

To determine

The value of H as F approaches zero.

(d)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of H is also approaches to zero as F approaches zero.

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for H in terms of F is,

H=F2(2L)(m2g2+F2)

In the above expression, the height of the ball is directly proportional to the square of the magnitude of force as the force increases then the height of the ball also increases but in the given case, the value of force is approach to zero then the height of the ball also approach to zero.

Conclusion:

Therefore, the value of H is also approaches to zero as F approaches zero.

(e)

To determine

The value of H as F approaches infinity.

(e)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of H as F approaches infinity is 1.6m .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for H in terms of F is,

H=F2(2L)(m2g2+F2)=2L(m2g2F2+1)

Substitute for F and 80.0cm for L in the above formula to find H .

H=2L(m2g2F2+1)=2(80.0cm×102m1cm)(m2g2()2+1)=1.6m(0+1)=1.6m

Conclusion:

Therefore, the value of H as F approaches infinity is 1.6m .

(f)

To determine

The equilibrium height of the ball as a function of F .

(f)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The equilibrium height of the ball as a function of F is (0.08m0.23522.93+F2)

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 8, Problem 8.82CP , additional homework tip  2

Figure II

The diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 8, Problem 8.82CP , additional homework tip  3

Figure III

From the figure the equilibrium height of the ball is,

Heq=LLcosθ

Here,

L is the length of the string.

θ is the angle between the string and the vertical.

From the figure II,

cosθ=mg(mg)2+F2

Substitute mg(mg)2+F2 for cosθ in formula (1) to find Heq .

Heq=LLcosθ=LL(mg(mg)2+F2)=L(1(11+(Fmg)2))

Substitute 300g for m , 80.0cm for L and 9.8m/s2 for g in the above formula to find Heq .

Heq=(80.0cm×102m1cm)(1(11+(F(300g×103kg1g)(9.8m/s2))2))=(80.0cm×102m1cm)(111+F28.64N2)

Conclusion:

Therefore, the equilibrium height of the ball as a function of F is (80.0cm×102m1cm)(111+F28.64N2) .

(g)

To determine

The value of equilibrium height of the ball H for F=10N .

(g)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of equilibrium height of the ball H for F=10N is 0.5744m .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for equivalent height of the ball in terms of F is,

Heq=LL(mg(mg)2+F2)

Substitute 10N for F , 300g for m , 9.8m/s2 for g , 80.0cm for L in the above formula to find Heq .

Heq=(80.0cm×102m1cm)(1((300g×103kg1g)(9.8m/s2)((300g×103kg1g)(9.8m/s2))2+(10N)2))=(80.0cm×102m1cm)(12.940N(2.940N)2+(10N)2)=(0.8m)(10.282N)=0.5744m

Conclusion:

Therefore, the value of equilibrium height of the ball H for F=10N is 0.5744m .

(h)

To determine

The value of equilibrium height of the ball H as F approaches infinity.

(h)

Expert Solution
Check Mark

Answer to Problem 8.82CP

The value of equilibrium height of the ball H as F approaches infinity is 0.8m .

Explanation of Solution

Given info: The length of the string is 80.0cm , mass of the ball is 300g .

From part (a), the expression for equivalent height of the ball in terms of F is,

Heq=LL(mg(mg)2+F2)

Substitute for F and 80.0cm for L in the above formula to find Heq .

Heq=(80.0cm×102m1cm)(1(mg(mg)2+()2))=(80.0cm×102m1cm)(1mg)=(80.0cm×102m1cm)(10)=0.8m

Conclusion:

Therefore, the value of equilibrium height of the ball H as F approaches infinity is 0.8m .

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
63. A 10.0-kg block is released from rest at point in Fig- Mure P8.63. The track is frictionless except for the por- tion between points and ©, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2 250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points and ©. 3.00 m 6.00 m Figure P8.63
To form a pendulum, a 0.092 kg ball is attached to one end of a rod of length 0.62 m and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?
40. A pendulum, comprising a light string of length L and a S small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.40). (a) Show that if the sphere is released from a height below that of the peg, it will return to this height after the string strikes the peg. (b) Show that if the pendu- Pege lum is released from rest at the hor- izontal position (0 = 90°) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5. Figure P8.40

Chapter 8 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning