Tougher Practice Problems for Physics 1020, ex3, Fall 2011, Solutions
1.
1. Ans
On the planet Feldspar, a 3.00 kg ball dropped from rest from a height of 70.0 meters will take
4.00 seconds to reach the ground. What is the weight of the ball on the planet Feldspar?
Ignore air resistance.
H = ½ g t2 => g = 2 × 70 ÷ ( 42) = 8.75 m/s2.
Weight = Mg = 3 × 8.75 = 26.25 N Ans.
2.
You wish to move a 20 kg object across a rough floor on planet Earth. In order to reduce the
amount of friction that you must overcome should you (a) pull the object with a force of 50.0 N
at an angle of 35° above the horizontal, or (b) push the object with a force of 40.0 N at an angle
of 25° below the horizontal? Answer “pull” or “push” NOT “a” or “b”. Hint: remember
that this is a Fill In question!
2. Ans.
Pulling upward at an angle above the horizontal deceases the support (normal) force,
thereby decreasing friction. So it is easier to pull than to push. Ans.
3.
A 30.0 kg object is initially at rest along a frictionless, horizontal surface. It is pulled with a
horizontal force of 75.0 N. How far does the object move in 4.00 seconds?
3.Ans
a = Fapp ÷ M = 75 ÷ 30 = 2.5 m/s2. Then x = ½ a t2 = ½ × 2.5 × 42 = 20 m. Ans.
4.
The coefficients of static and kinetic friction between an 80 kg object and a horizontal surface
are μs = 0.300 and μk = 0.150. How hard must I pull (horizontally) on the object in order to
get it moving? Once it is moving, how hard must I pull (horizontally) on the object to have
it accelerate at 2.00 m/s2?
4.Ans
In all parts, the weight = Mg = 80 × 9.8 = 784 N. Since we don’t want any vertical motion,
and the Fapp is horizontal (no vertical component), ΣFvert = 0 => η – Mg = 0 => η = 784 N.
I must pull horizontally hard enough to overcome static friction in order to get it moving. Since
fs,max = μs × η, we have fs,max = 0.300 × 784 = 235.2 N. So we must pull with at least 235.2 N
in order to get the object moving. Ans.
Newton’s Second law tells us that ΣFhoriz = Fapp – fk = M a. We are given that the acceleration
is 2 m/s2. Also, since fk = μk × η , we have since fk = 0.150 × 784 = 117.6 N. Substituting, we
get Fapp – fk = M a => Fapp – 117.6 = ( 80 × 2 ) => Fapp = 277.6 N. Ans.
5.
A 20 kg object is initially at rest on a horizontal table. It is acted upon by a horizontal force
Fapp. The coefficients of static and kinetic friction between the object and the table are μs = 0.3
and μk = 0.2.
(a) What is the minimum Fapp that will allow the object to start moving?
(b) Once the object is moving, what is Fapp if we observe that the object is moving at constant
speed?
(c) What is Fapp if we observe that the object moves 40 meters in 4 seconds?
(d) If Fapp = 80 N, how fast is the object moving after 6 seconds?
5. Ans
In all parts, the weight = Mg = 20 × 9.8 = 196 N. Since we don’t want any vertical motion,
and the Fapp is horizontal (no vertical component), ΣFvert = 0 => η – Mg = 0 => η = 196 N.
a)
I must pull horizontally hard enough to overcome static friction in order to get it moving. Since
fs,max = μs × η, we have fs,max = 0.300 × 196 = 58.8 N. So we must pull with at least 58.8 N
in order to get the object moving. Ans.
In (b) and (c) the object is moving, so we are dealing with kinetic friction. Since fk = μk × η ,
we have since fk = 0.200 × 196 = 39.2 N
Newton’s second law applied to the horizontal forces gives us, in all remaining parts,
ΣFhoriz = Fapp – fk = M a ; so Fapp – 39.2 = 20 a => Fapp = 39.2 + ( 20 a )
b)
We want a constant speed, so a = 0, and Fapp = 39.2 + ( 20 × 0 ) = 39.2 N. Ans.
c)
x = ½ a t2 => 40 = ½ × a × 42 => a = 5 m/s2, and Fapp = 39.2 + ( 20 × 5 ) = 139.2 N. Ans.
d)
We want a for Fapp = 80 N => 80 – 39.2 = 20 a => a = 2.04 m/s2. Then
v = v0 + a t = 0 + ( 2.04 × 6 ) = 12.24 m/s. Ans.
6.
I weigh 210 lbs.  My mass is 95 kg. The mass of the earth is 5.98 × 1024 kg. What is the
gravitational force of me on the earth? And yes, I know that I gave you my weight in pounds,
not Newtons. So give me the answer in pounds as well!
6.Ans
From Newton’s third law, the force of gravity of me on the earth is equal (and opposite) to
the force of gravity of the earth on me. Since the force of gravity of the earth on me is my
weight, the force of gravity of me on the earth is also equal to my weight, 210 lbs. Ans.
7.
Support
3
0
k
g
Mass of block = 30 kg
Fapp
friction
Weight
In the figure drawn above, the mass of the block is 30 kg.
a) It is determined that an applied force of Fapp = 80 N is necessary to get the block moving. Find the
coefficient of static friction.
b) Once it is actually in motion, a smaller force will generally keep it in motion. It is determined that
an applied force of Fapp = 50 N will keep the object moving at a constant velocity. Find the
coefficient of kinetic friction.
c) Given the coefficient of kinetic friction calculated in (b), what would be the acceleration of the
block if the applied force was increased back up to Fapp = 80 N?
7. Ans
In all parts, the weight = Mg = 30 × 9.8 = 294 N. Since we don’t want any vertical motion,
and the Fapp is horizontal (no vertical component), ΣFvert = 0 => η – Mg = 0 => η = 294 N.
a)
We need Fapp > fs,max., or since we know that Fapp = 80 N gets us moving, 80 = fs,max. Then,
since fs,max = μs × η, we have 80 = μs × 294 => μs = 0.272. Ans.
In (b) and (c) the object is moving, so we are dealing with kinetic friction. Newton’s second
law applied to the horizontal forces gives us
ΣFhoriz = Fapp – fk = M a ; and since fk = μk × η , we get Fapp – ( μk × 294 ) = 30 × a.
b)
We are told that for Fapp = 50 N, we move at constant velocity, i.e. a = 0. So
50 – ( μk × 294 ) = 30 × 0 => μk = 0.170 Ans.
c)
We want a for Fapp = 80 N and μk = 0.170.
80 – ( 0.170 × 294) = 30 × a => a = 1.00 m/s2. Ans.
In problems 8 through 10, I am standing on a scale on top of tower that is firmly anchored onto Earth’s
surface. My mass is 80 kg (I wish). All objects are at rest. I’ll put a drawing on the board on
Tuesday, November 22.
8.
What does the scale read?
a) zero, by Newton’s Second Law.
b) 80 N, by Newton’s Third Law.
9.
What is the force of me on the tower? Think carefully!!, especially about what a free–body
diagram of the tower would look like!
a) zero
10.
c) 784 N, by Newton’s Second Law
d) 784 N, by Newton’s Third Law
b) 80 N
c) 80 kg
d) 784 N
What is the force of me on the earth?
a) zero, by Newton’s Second Law.
b) 80 N, by Newton’s Third Law.
c) 784 N, by Newton’s Second Law
d) 784 N, by Newton’s Third Law
The answers are
8. The sum of the forces on me is zero since I am rest. The forces on me are gravity down and the
force from the scale up. The force of gravity down is my weight = 80 × 9.8 = 784 N, down. In order
for the sum of the forces on me to be zero (Newton’s second law) the force of the scale on me must
also be 784 N. So the answer is C.
9. I am not touching the tower, so I do not exert a force on the tower. Therefore, the answer is A.
10. The force of me on the earth is the reaction force, in the sense of Newton’s Third Law, to the
force of the earth on me. The force of earth on me is due to gravity, and is therefore my weight, which
we calculated in #8 to be 784 N. So the answer is 784, by Newton’s Third Law, D. Ans.