Download UNIT 2 REVIEW SHEET Answers sp 10

UNIT 2 REVIEW SHEET and ANSWERS
(CHAPTER 4)
1. If an object here on Earth has a mass of 120 kg, on Jupiter where
acceleration due to gravity is 25 m/s2, that same object would weigh
what on Earth and on Jupiter? On Jupiter it would have a mass of ?
Weight on Earth Fw = mg 120(10) =
Weight on Jupiter Fw = mg 120 (25)=
The object would have the same mass on Jupiter as it does on Earth.
Mass doesn’t change.
2. When a table cloth is pulled out quickly from underneath plates, this
can be done successfully because the plates have Inertia? This is an
example of Newton’s First law.
3. What is friction? A force that opposes motion
4. Acceleration is the result of Force.
5. At terminal velocity acceleration is Zero!.
6. What is Newton’s law of inertia? The first law, which says that an
object at rest, or an object in motion, will stay at rest, or continue its
motion until acted upon by an unbalanced force.
7. If a ball rolls down an incline, how far does it roll up an incline
opposite the first one? With friction, the ball will not roll back up the
incline the same distance. Without friction it will roll an equal
distance back up the incline
8. If a 50 N force acts towards the west and a 20 N force acts towards the
east, what is the net force?
Fnet = Right – Left = 20-50 = -30 N or 30 N left
9. A 100 N force acts northward, a 40 N force acts southward and a 30 N
force acts towards the east. Fnet = Top – Bottom = 100-40 = 60 N
North. There is a force now going North and East. Use the
Pythagorean Theorem when you have vectors at right angles to each
other. X2 + Y2 = R2 What is the resulting net force? 302 + 602 = R2
10.If you push a 50 kg object with 100 N of force, how much will it
accelerate? Fnet = ma = 50(100) = 5000 N
11.Forces always occur in pairs.
12.If a golf club hits a golf ball, what is the reaction to this force? The
golf ball hits the golf club
13.What is Newton’s third law? For every action or force there is an
equal and opposite reaction or force.
14.A Mack truck hits a VW bug. Compare the forces of impact. The
force on impact is the same for both according to Newton’s 3rd law.
Which one will move more after impact? The bug experiences a
greater acceleration (in the negative direction) b/c it stops but the
Mack truck keeps on moving.
15.A golf ball leaves the tee at a speed of 75 m/s. If the ball has a mass
of 0.05 kg and gains this speed in 0.02 seconds, what is the average
force of impact of the club on the ball? Fnet = ma = (.05)(75-0/ .02)
=187.5
16.If an object weighs 600 N what is its mass? Fw = mg
600 =m(10) = 60 kg
17.How much does a 15 kg object weigh? Fw = mg 15(10) = 150 N
18.If you push on a box with 40 N of force, and the box has a mass of 20
kg and there is a 20 N friction force, how fast will the box accelerate?
Fnet = Right – Left = 40-20 = 20N Fnet = ma
20 = m(20)
1kg = m
19.What is terminal velocity or speed? A constant speed/velocity that an
object reaches when its weight is equal to the air underneath it.
20.At terminal velocity, what is an object’s acceleration? Zero
21.At terminal velocity, what is air resistance equal to? The weight of
the object
22.Without friction, once an object is set into motion, does it require
force to keep it moving? No
23.What is Newton’s second law? F = ma An object’s mass will resist its
acceleration as a force is applied.
24.Forces are VECTOR quantities.
25.Can an object be moving and have zero acceleration? Explain your
answer. Yes, if it has a constant speed.
26.If you stand on a scale, does your weight change if you lift one foot
off the scale? No Why or why not? Weight is dependent on the
downward force of gravity. Gravity doesn’t change just b/c you lifted
one foot.
27.What is the relationship between mass and terminal velocity. The
greater the mass, the greater the terminal velocity, direct relationship.
28.What does equilibrium mean? All the forces are equal on the object.
29.Why do we wear seatbelts in a car? To help us resist the change in
motion that the car gives us.
30. If you push with 50 N of force on a 20 kg crate you encounter 10 N
of friction. Fnet = Right – Left = 50- 10 = 40 N How fast will the
crate accelerate. Fnet = ma 40 = 20(a) = 40/20 = 2.0 m/s
31. Why is F = ma and Fw = mg virtually the same equation.? The only
difference in these two equations is that I am specifying that “a”,
which is acceleration, has to be the value of gravity when calculating
weight.
32.Mira received a 235 N sled for her birthday. She takes the sled out to
a flat snowy field. When she pushes it with a 45 N horizontal force, it
slides with a constant speed. What is the coefficient of kinetic friction
between the sled and the snow? μ = Ff/FN Since the sled is moving at
a constant speed the pushing force is equal to the frictional force. The
normal force is equal to the weight force. It gives us the weight of the
sled (235 N) and it tells us the pushing force is 45 N.
μ = Ff/FN = 45/235 = .191
33.A worker has to move a 17.0 kg crate along a flat floor in a
warehouse. The coefficient of kinetic friction between the floor and
crate is 0.214. The worker pulls horizontally on a rope attached to the
crate, with a 49 N force. What is the acceleration of the crate? Since
the crate is accelerating you know there is a net force on the crate.
You have to figure out the Ff first and then use this value to find the
net force on the crate. You also have to calculate the weight of the
crate using the given mass. The normal force is equal to the Weight
force. μ = Ff/FN = .214 =Ff/170 = 36.38 N = Ff Find the net Force
now. Fnet = Right –Left Fnet = 49 – 36.38 = 12.62 N Now find the
acceleration using the net force and the original mass given in the
problem. Fnet = ma = 12.62 = 17a
.74 m/s2 =a
34.What is the force applied to a 1500 kg car moving at a constant
velocity on a horizontal road when the coefficient of kinetic friction is
0.04. Since the car is moving at a constant velocity the Ff = Fp.
Calculate the Ff using the equation μ = Ff/FN = .04 =Ff/15000.
Notice that I turned the mass of 1500 into a weight by multiplying it
by 10. Ff = 15000(.04) = 600 N
Find the effect on the gravitational force if …..
For all of these problems you use the ones rule.
35. both masses are halved. F = Gm1m2/r2
F = (1)(1/2)(1/2)/12 = ¼ /1 = ¼
36. one mass is doubled and the distance is halved
F =(1)(2)(1)/ (½)2 = 2/¼ Use the keep it change it flip it rule. Keep the
2, change the division to multiplication and flip the denominator from ¼
to 4 = 8 x the original gravitational force.
37. one mass is cut to ¼ and the distance is tripled = .027 x
38. one mass is cut to ½ and the distance is doubled =.125 x
39. both masses are halved and the distance is quadrupled =1/64 x
40.one mass is doubled, one mass is cut to 1/4 and the distance is
doubled = 1/8 x
41.one mass is tripled and the other mass is halved =3/2 =1.5 x
42.the distance is cut to 1/10 = 100 x
43.one mass is quadrupled and the distance is cut to 1/10 = 400 x