Download Two Cars on a Curving Road

skiladæmi 4
Due: 11:59pm on Wednesday, September 30, 2015
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Two Cars on a Curving Road
Part A
A small car of mass m and a large car of mass 4m drive along a highway at constant speed. They approach a curve
of radius R. Both cars maintain the same acceleration a as they travel around the curve. How does the speed of the
small car v S compare to the speed of the large car v L as they round the curve?
Hint 1. How to approach the problem
To solve this problem use proportional reasoning.
Find the simplest equation that contains the variables and other known quantities from the
problem.
Write this equation twice: once for the car of mass m and speed v S and again for the car of mass 4m and speed v L .
Write each equation so that all the constants are on one side, cancel any quantities that appear on
both sides and keep the variables on the other side. In this problem the variable is v so write your
equations in the form v = ….
Finally, compare the two cases presented in the problem. For this question you should find the
ratio v S /v L .
Hint 2. Determine which equation to use
In this problem you are told that the cars need to experience the same acceleration as they drive around a curve
of radius R. You are also given information that relates the masses of the cars. Which equation is best to use to
find out information about the speed each car can travel?
ANSWER:
a
2
c
=
⃗ F net
F
c
v
R
= ma⃗ 2
=
mv
R
Hint 3. Relating the speeds of the cars
You just established the relationship
.
Since a and R are the same for both cars we can conclude that
aR = v
v
2
L
= v
2
2
S
.
Use this equation with what is known about the masses to find the relationship between v 2S and v 2L .
ANSWER:
v
=
v
=
vS
= vL
vS
= 2v L
vS
= 4v L
S
S
1
4
1
2
vL
vL
Correct
Now apply this method for using proportional reasoning to the next problem.
Part B
Now assume that two identical cars of mass m drive along a highway. One car approaches a curve of radius 2R at
speed v . The second car approaches a curve of radius 6R at a speed of 3v. How does the magnitude F1 of the net
force exerted on the first car compare to the magnitude F2 of the net force exerted on the second car?
Hint 1. Determine the equation to use for the comparison
Which of the following equations makes the correct comparison between car 1 and car 2?
Hint 1. How to approach the problem
To solve this problem use proportional reasoning to find a relation among F , v , and R.
Find the simplest equation that contains these variables and other known quantities from
the problem.
Write this equation twice: once to describe F1 , v 1 , and R1 and again to relate F2 , v 2 , and R2 .
Write each equation so that all the constants are on one side and the variables are on the
other. In this problem the variable is F so write your equations in the form F = ….
Finally, compare the two cases presented in the problem. For this question you should find
the ratio F1 /F2 .
Hint 2. Equation for the net force acting on each car
The net force acting on each car as it travels around the highway curves is given by
2
Fc =
mv
R
,
where m is the mass of the car, R is the radius of the curve, and v is the speed of the car.
ANSWER:
F
1
v2
1
R1
F
1
=
R1
2
v
F2 v2
=
1
m
1
R2
1
2
v
2
F2 R2
2
v
2
v2
1
R1
m
2
R2
=
=
m 2 v2
2
R2
m 2 R1
2
v
1
ANSWER:
F
=
F
=
F1
= F2
F1
= 3F 2
F1
= 27F 2
1
1
1
3
3
4
F2
F2
Correct
Being able to use proportional reasoning is a useful skill to have. The more you practice, the better you will
become at it.
Problem 5.76
Block A in the figure weighs 61.6 N . The coefficient of static
friction between the block and the surface on which it rests is
0.26. The weight w is 12.2 N and the system is in equilibrium.
Part A
Find the friction force exerted on block A.
ANSWER:
f
= 12.2 N Correct
Part B
Find the maximum weight for which the system will remain in equilibrium.
ANSWER:
wmax
= 16.0 N Correct
Problem 5.89
Block A in the figure has a mass of 4.50 kg , and block B has
mass 13.0 kg . The coefficient of kinetic friction between block B
and the horizontal surface is 0.15.
Part A
What is the mass of block C if block B is moving to the right and speeding up with an acceleration 1.90 m/s2 ?
ANSWER:
mC
= 12.2 kg Correct
Part B
What is the tension in each cord when block B has this acceleration?
ANSWER:
TAB
= 52.7 N Correct
Part C
ANSWER:
TBC
= 96.5 N Correct
± The Work Done in Pulling a Supertanker
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.40×106 N , one at an angle 11.0 ∘ west of
north, and the other at an angle 11.0 ∘ east of north, as they pull the tanker a distance 0.640 km toward the north.
Part A
What is the total work done by the two tugboats on the supertanker?
Express your answer in joules, to three significant figures.
Hint 1. How to approach the problem
There are two ways to calculate the total work done on an object when several forces act on it. You can
compute the quantities of work done on the object by each force and then add them together. Alternatively, you
can compute the work done on the object by the net force acting on it. The hints that follow are meant to help
you to calculate the total work using the first method.
Hint 2. Find the work done by one tugboat
What is the work done on the tanker by the tugboat that exerts a force in the direction west of north?
Express your answer in joules, to three significant figures.
Hint 1. The definition of work
⃗ The work W done by a constant force F acting on an object that undergoes a straight­line displacement d
⃗ is given by the formula
,
W = F d cos ϕ
where ϕ is the angle between the direction of the force and the direction of displacement.
ANSWER:
8.80×108 J ANSWER:
1.76×109 J Correct
± All Work and No Play
Learning Goal:
To be able to calculate work done by a constant force directed at different angles relative to displacement
If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done
on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the
force can be calculated as
⃗ ∣
⃗ ∣
⃗ ⃗ ∣
∣
⃗ cos θ
W = F ⋅ s ⃗ = F ∣
∣s ∣
∣
∣
∣
,
⃗ ⃗ ⃗ where W is the work done by force F on the object that undergoes displacement s directed at angle θ relative to F .
Note that depending on the value of cos θ, the work done can be positive, negative, or zero.
In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is
related to the accompanying figure.
Part A
What can be said about the sign of the work done by the force F 1⃗ ?
ANSWER:
It is positive.
It is negative.
It is zero.
There is not enough information to answer the question.
Correct
When θ
= 90
∘
, the cosine of θ is zero, and therefore the work done is zero.
Part B
What can be said about the work done by force F 2⃗ ?
ANSWER:
It is positive.
It is negative.
It is zero.
Correct
When 0∘
< θ < 90
∘
, cos θ is positive, and so the work done is positive.
Part C
⃗ The work done by force F 3⃗ is
ANSWER:
positive
negative
zero
Correct
When 90 ∘
∘
< θ < 180
, cos θ is negative, and so the work done is negative.
Part D
The work done by force F 4⃗ is
ANSWER:
positive
negative
zero
Correct
Part E
The work done by force F 5⃗ is
ANSWER:
positive
negative
zero
Correct
Part F
The work done by force F 6⃗ is
ANSWER:
positive
negative
zero
Correct
Part G
The work done by force F 7⃗ is
ANSWER:
positive
negative
zero
Correct
In the next series of questions, you will use the formula W
⃗ ⃗ ∣
∣
⃗ cos θ
= F ⋅ s ⃗ = F ∣
∣s ∣
∣
∣
∣
to calculate the work done by various forces on an object that moves 160 meters to the right.
Part H
Find the work W done by the 18­newton force.
Use two significant figures in your answer. Express your answer in joules.
ANSWER:
W
= 2900 J Correct
Part I
Find the work W done by the 30­newton force.
Use two significant figures in your answer. Express your answer in joules.
ANSWER:
W
= 4200 J Correct
Part J
Find the work W done by the 12­newton force.
Use two significant figures in your answer. Express your answer in joules.
ANSWER:
W
= ­1900 J Correct
Part K
Find the work W done by the 15­newton force.
Use two significant figures in your answer. Express your answer in joules.
ANSWER:
W
= ­1800 J Correct
Exercise 6.42
A block of ice of mass 4.50 kg is placed against a horizontal spring that has force constant k = 190 N/m and is
compressed a distance 2.10×10−2 m . The spring is released and accelerates the block along a horizontal surface. You can
ignore friction and the mass of the spring.
Part A
Calculate the work done on the block by the spring during the motion of the block from its initial position to where the
spring has returned to its uncompressed length.
Express your answer using two significant figures.
ANSWER:
W
= 4.2×10−2 J Correct
Part B
What is the speed of the block after it leaves the spring?
Express your answer using two significant figures.
ANSWER:
v
= 0.14 m/s Correct
Exercise 6.34
To stretch a spring 9.00 cm from its unstretched length, 19.0 J of work must be done.
Part A
What is the force constant of this spring?
ANSWER:
= 4690 N/m k
Correct
If you need to use the value of the spring constant 'k' in subsequent parts, please use the unrounded full precision
value and not the one you submitted for this part rounded using three significant figures.
Part B
What magnitude force is needed to stretch the spring 9.00 cm from its unstretched length?
ANSWER:
F
= 422 N Correct
Part C
How much work must be done to compress this spring 4.00 cm from its unstretched length?
ANSWER:
W
= 3.75 J Correct
Part D
What force is needed to stretch it this distance?
ANSWER:
F
= 188 N Correct
Score Summary:
Your score on this assignment is 101%.
You received 7.07 out of a possible total of 7 points.