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SOLUTIONS FOR HOMEWORK #5

Page 57 questions

Question 21:
We know from the text that weight and mass are related, but not equivalent concepts. The mass of an object is how much matter it contains, its weight is a measure of the gravitational force of attraction between the Earth and the object. As the book notes on page 50:

W = mg

where W is weight, m is mass, and g is the acceleration due to gravity. In this case:
W= mg

W = 50 kg x 10 m/sec/sec = 500 N

(as mentioned in class, we can always approximate the value of g to be 10 m/sec/sec for ease of calculations.

Question 25:

W=mg

Some simple division (divide each side of the equation by m yields:
m = W/g = 300 N/ 10 m/sec/sec = 30 kg

Question 26:
A person's head is non rigidly supported to the body by the neck. In a rear-end collision, the car is bumped forward, and the body (well supported by the upright portion of the seat) moves forward with the same speed as the car. However, the head has some inertia, and moves along at the original speed of the car. Hence, the head lags behind the body, and when the rest of the car catches up with the head, causes significant injury in the impact. The purpose of the headrest is to connect the head to the motion of the car, so there is no differential motion between head and body, or between head and car.

Question 27:
This is very similar to the question above. If there is little friction between the wagon and ball, the ball's motion is not tied to that of the wagon. So, the ball's inertia tends to keep it at rest; when the outside force pulls the wagon forward, the ball's inertia keeps it at rest (apart from some slight motion dependent on friction between the ball and bed of the wagon). So, with respect to the wagon, the ball appears to move backward; with respect to an observer on the sidewalk, it appears as is the ball is stationary and it is the wagon moving forward.

In part a), the tension in the upper string is greater because it is supporting the forces due both to the weight of the ball and the tension on the string. In this case, the lower string supports only the tension caused by the person's pull. In part b), where I am applying a sudden force to snap the string, the lower string is more likely to snap since it is trying to change the inertia of the ball in a short time, thus generating a large force on the lower string, and making it more likely to snap.

Page 72 Questions

Question 21:
This is an example of Newton's Second Law of Motion, given by:

F = ma

a) In this case, we are asked to find the acceleration, so we write:
a = F/m = 20 N/ 2 kg = 10 m/s/s

b) In this case, the opposing frictional force has a value of 4 N. We must be careful to realize that it is not the force of friction that is accelerating this object, rather, the force is the net force that is producing the acceleration.<
F = ma

becomes:
Pushing force - frictional force = 20 N - 4N = ma

or:
16 N = ma implies a = 16 N/2kg = 8 m/s/s

Question 23:
Again, we start with F = ma, but we know that we want the acceleration to be 1.8 times the acceleration due to gravity, so:

F = ma = 1.2 kg x 1.8 x 10 m/s/s = 21.6 N

In this problem, you needed to recognize that g has the value of 10 m/s/s.

Question 24:
When two quantities are proportional, we are saying that changing the value of one quantity will cause the other to change in a similar way. For instance, we can say that in Illinois state income tax paid and total income are proportions; if you earn twice as much in IL you will pay twice as much in state taxes, if you make 1/3 as much you will pay 1/3 as much in taxes. A statement of proportionality tells you how quantities vary with respect to each other, but does not allow you to calculate the EXACT value of either quantity.

Statements that are exact means that one side of the equation exactly equals the other side; in other words, if two quantities are identical, they will yield a value of 1 when divided.

Question 26:
All objects in motion in the Earth's gravitational field (neglecting any effects of air friction) are accelerated downward at the acceleration due to gravity. So at all points during the trip, the acceleration of this object is 10 m/s/s toward the center of the Earth.

To show this with equations, use:

a = F/m

remembering that in this case the force is the weight of the object (which is Mg), so:
a = F/m = Mg/M = g

Question 28:
This question requires you to recognize that the book is moving at constant speed. Objects moving with constant speed are not accelerating, and that means that there are no net forces acting on the book. Since we are told that there is a pushing force of 1 N acting on the book, this means that the force of friction acting in opposition to the push must be 1 N to produce a net force of zero, and allow the book to move at constant speed in a straight line.


Loyola University Chica
go David B. Slavsky
Loyola University Chicago
Cudahy Science Hall, Rm. 404
1032 W. Sheridan Rd.,
Chicago, IL 60660
Phone: 773-508-8352
and:
IES 310
phone: 773-508-2149
E-mail: dslavsk@luc.edu

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