Download Newton`s Laws Problem Solving Worksheet Name: Use g = 10.0 m

Newton’s Laws Problem Solving Worksheet
Name:
_______________________________
Use g = 10.0 m/s2
1. When a karate expert strikes a wooden block, the hand undergoes an acceleration of -6500
m/s2. Medical data indicates the mass of the forearm and hand to be about 0.70 kg. What is
the force exerted on the hand by the blocks? What is its direction?
Newton’s third: Fbloock on hand = Fhand on block
Fblock on hand = ma = 0.7 * 6500 m/s2 = -4550N in direction of movement
Fhand on block = 4550 in direction opposite movement = 4600N
2. When you drop a 0.40 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 toward
the Earth’s surface. According to Newton’s third law, the apple must exert an equal and
opposite force on the Earth. What force does the Earth feel? If the mass of the Earth is 5.98 *
1024 kg, what’s the magnitude of Earth’s acceleration?
Apple.
Earth Gravity: 9.81m/s^2
Apple gravity: unknown
Apple-mass: 0.4kg,
Earth
G=?
Earth –mass:5.98 ∗ 1024 𝑘𝑔,
G= 9.81𝑚/𝑠 2
Apply Theory/ Plan: Since this question is asking about acceleration I will be using𝐹 = 𝑚𝑎. I know that
Newton’s third law says that Force of two objects are same. So𝐹𝑎𝑝𝑝𝑙𝑒 = 𝐹𝑒𝑎𝑟𝑡ℎ , and I can substitute the
formula to0.4 × 𝑔𝑒𝑎𝑟𝑡ℎ = 5.98 × 1024 × 𝑔𝑎𝑝𝑝𝑙𝑒 . The reason that I am multiplying mass of apple with
gravity of earth is because; earth is pulling the apple down. And apple also pulls the earth but the mass of
the earth is so great that it gets pulled really short in length.
Formulas and solve:
𝐹 = 𝑚𝑎
𝐹𝑎𝑝𝑝𝑙𝑒 = 𝐹𝐸𝑎𝑟𝑡ℎ
0.4𝑘𝑔 ×
9.81𝑚
= (5.98 × 1024 𝑘𝑔) × 𝑔𝑎𝑝𝑝𝑙𝑒
𝑠2
9.81𝑚
𝑠2
= 𝑔𝑎𝑝𝑝𝑙𝑒
24
(5.98 × 10 𝑘𝑔)
0.4𝑘𝑔 ×
𝑔𝑎𝑝𝑝𝑙𝑒 = 6.25 × 10−25 𝑚/𝑠 2
Check: This answer makes sense because the gravity of apple is smaller compare to gravity of earth.
3. A 60 kg boy and a 40 kg girl use an elastic rope while engaged in a tug of war on an icy
frictionless surface. If the acceleration of the girl toward the boy is 3 m/s2 , determine the
magnitude of the acceleration of the boy toward the girl.
Newton’s Third Law: Fboy on girl = Fgirl on boy
Only forces acting on each kid in the horizontal direction is the Tension, so set them each equal
to ma
60 * aboy= 40 * 3
A = 2 m/s2
4. After a day of testing race cars, you decide to take your own 1550 kg car onto the test track.
While moving down the track at 10 m/s, you suddenly accelerate to 30 m/s in 10 sec. What is
the average net force that you have applied to the car in the 10 second interval?
N
Diagram and Givens:
u = 10 m/s, v = 30 m/s, t = 10 s, m = 1550kg
Fapplied
W
Apply Theory/Plan: I know that Newton’s First Law says that unless an unbalanced force acts on an
object, there will be no chance in motion. Since I am starting with a velocity of 10 m/s and ending with a
velocity of 30 m/s, I know that there was a change in motion and therefore according to Newton’s
Second Law, there must have been an acceleration. I know that whatever net force acts, that Fnet =
m*a. Since I am given the time over which I accelerate, I can find the acceleration of the car. Since I
know the mass of the car, then I can plug in to find the force that is needed to make this change in
motion. While a full force diagram is shown above, the Normal and the Weight cancel (which makes
sense since the car doesn’t leave the ground) so we do not need this part in order to solve our problem,
since we are only asked to find the force in the “x” direction.
Formulas and Solve:
𝑎=
𝑎=
𝑣−𝑢
𝑡
and F = ma
30𝑚/𝑠 − 10𝑚/𝑠 20𝑚/𝑠
=
= 2𝑚/𝑠 2
10𝑠
10𝑠
F = 1550kg* 2m/s2= 3100 kg m/s2 = 3100 Newtons
Check: This answer makes sense because a very massive car would take quite a bit of force to
accelerate at that rate. The units work out in each step, ending with a force in Newtons. There is no
other way to solve this problem.
5. A force of -9000 N is used to stop a 1500 kg car traveling at 20 m/s. What braking distance is
needed to bring the car to a halt?
Diagrams and givens:
Apply Theory/Plan: Newton’s first law says that an object in motion tends to stay in motion and an
object at rest tends to stay at rest unless acted upon by an outside force. The car has an initial velocity
of 20 m∙s-1 and the car ended up with a final velocity of 0 m∙s-1, there was a change in motion, which
means that we can use Newton’s second law of motion to show that there must have been a
deceleration. First of all we do not need to worry about the normal and weight in the force diagram
shown above because they cancel (the car doesn’t leave the ground, and the car doesn’t go through the
ground). First I need to find the deceleration, I do this by using the equation F=m∙a. We are given the
mass of the car and the force it took to make the car stop, so I can plug those values into the equation
and get the deceleration. Next we need to find the braking distance; to do this I need to use the
equation v2=u2+2as. Initial velocity is given, I know the final velocity, and I have found the deceleration.
Now I plug in the values and solve for “s” to get the total braking distance.
Formulas and Solve:
Check: the answer that I got does make sense because since the car is quite heavy and is traveling at 20
m∙s-1 (72 km∙h-1) it would require quite some distance to for the car to come to a full stop.
#5 Method 2:
Diagrams/given:
u = 20m/s
v= 0m/s
F=-9000N
s= ?
m = 1500kg
Apply theory/plan: Newton's second law of motion explains how an object will change velocity if it is
pushed or pulled upon. Since our car is already going 20m/s, that will be stopped by a Force (-9000N).
this tells us that the car will stop to 0m/s. since it has to do with deceleration/ negative acceleration, I
know it’s the second law of motion. That means I can use the F=ma formula. (Force = mass X
acceleration). So first I have to find the acceleration/deceleration. I will use my f=ma formula. I plug in
for “F” -9000N and for “m” 1500kg, and find “a”, which is -6m/s2. Now that I have the deceleration, I
can use my old formula “v2 = u2 + 2as”. I plug in my values, and what I get for “s” is 33.3metres.
Solving:
Formulas:
F=ma
v2 = u2 + 2as
work: -9000=1500a
a=-6m/s
0=202+2(-6)s
0=400-12s
-400=-12s
s=33.3m
Check: this answer seems pretty realistic, because it would take some time for a car to stop, and the force
o 9000N is pretty strong
6. Draw an accurate free body diagram of the forces on the block below. Determine the value of
each force you draw.
N = 15 N up
W = 15 N down
7. A car is traveling forward with a constant velocity. The total weight of the car and
passengers is 1200 N. The resultant force on the car must be
a. Greater than 1200 N
b. 1200 N
c. Between 1200 N and zero.
d. Zero.
8. You are driving a 2500 kg car at a constant speed of 14 m/s along an icy, but straight and
level road. You slam on the brakes and the wheels lock sending you sliding and skidding to a
halt in a distance of 25 m. What is the force of sliding friction between your tires and the icy
road?
V2 = u2 +2as
V=0
U = 14 m/s
S = 25 m
Mass = 2500 kg.
If I know the acceleration and the mass, then I can find the Force that needs to act on it in order
to cause that acceleration with F = ma. Using the formula above I can solve for acceleration: a =
-3.92 m/s2
Ffriction = ma = 2500kg * (-3.92 m/s2) = 9800 N
7.
9. Draw an accurate free body diagram of the forces on the still block below. Determine the
value of each force you draw.
N = 15 cos 25 = 13.6N
F = 15 sin 25 = 6.34 N
N = 14 N f = 6.3N
10. Several forces act on a particle. Find the resultant horizontal and vertical forces. Find the
magnitude and direction of the net force.
Diagrams and Givens:
 Fx  4.3 N
 Fx  7.9 N
 Fy  8.5 N
 Fy  3.5 N
Apply Theory/Plan: Newton’s second law states that “when an external unbalanced force acts on an
object the object will accelerate in the direction of the force.” This law is perfect to explain how this
particle moves, since both the x-axis and y-axis forces are unbalanced, this particle will move in the way
the stronger particles, in this case the negative x and the positive y. We simply need to find the
differences for the horizontal and vertical forces, then we must discover the theta which is found using
the final amount of the x and y forces, then to find the net force we also use the final amount of the x
and y forces.
Formulas and Solution:
Formulas needed:
Fnet  ( Fx ) 2  ( Fy ) 2
  tan 1 (
Fx
)
Fy
Answer: Force of 6N at 53° from the negative x-axis or 127° from the positive x-axis.
6N
53°
127°
Solution:
Fx  4.3  7.9  3.6 N
Fy  8.5  3.7  4.8 N
Fnet  (3.6) 2  (4.8) 2
Fnet  36
Fnet  6 N
  tan 1 (
4.8
)
3.6
  53
Check: This answer makes sense, because according the calculations a force of 6N can pull this particle at the direction
provided. Also the degree the particle moves seem correct.
11. Two blocks, one 10 kg and the other 20 kg, on a smooth horizontal surface are tied together. They are
accelerated by a horizontal force of 30 N as shown below. If friction effects are negligible, what is the
tension in the connecting rope?
30 n
10kg 
 20kg 
tension
f  ma
30  30a
a  30 / 30
1 a
a m/ f
t  ma
t  10 x1
t  10
12. D. Zero (again)
13. A traffic light weighing 100.0 N hangs from a cable tied to two other cables fastened to a support as shown
below. The upper cables make angles of 37º and 53º degrees with the horizontal. Find the tension in the
three cables.
Method 1:
T1y = T1 sin 37
T1x = T1cos 37
T1y +T2y –T3 = 0
T2y = T2sin 53
T1x = T2x
T2x = T2cos 53
T3 = 100N
Substitute: T1 sin 37 + T2sin 53 = 100N (1)
T1cos 37 = T2cos 53 (2)
Solve for T1:
T1 = T2(cos 53/cos 37) ( 2 rearranged)
Substitute (2 rearranged) into (1)
T2(cos 53/cos 37) sin 37 + T2sin 53 = 100N
Solve for T2
T2 = 79.8 = 80 N
Plug back in to solve for T1
T1 = 60.41 N = 60 N
Method 2: This method works because 37 + 53 = 90, so the angle formed between T1 and T2 is 90. However, it will not
work for angles that change. See method above for method that works for all angles.
Diagram and Givens:
Apply Theory/Plan:
The mass is hanging motionless and we know its weight. Therefore, we can easily find out the tension of T3
which is the same as the weight. After that, we can rearrange the tensions so they form a right triangle. From then, we
can easily find T1 and T2 using trigonometry, the calculated T3 and the angles given
Formulas and Solve:
Check:
These values make sense because logically, the sum of the tensions of T1 and T2 should be larger than T3
(although T3 by itself should be greater than either of the other two) as the two ropes were at an angle with the third,
vertical one.
14. When two unequal masses are hung vertically over a light, frictionless pulley as
shown , the arrangement is called Atwood’s machine. This device is sometimes
used in the laboratory to measure the acceleration of gravity. Determine the
acceleration of the two masses in terms of m1, m2 and g.
□1: _of Body1
Diagram and Givens:
□2: _of Body 2
All the forces:
T
T
M: Mass
g: gravity
M1
M1
M2
W: Weight
M2
T: Tension
W1=M1*g
a: Acceleration
W2=M2*g
Fnet in M1 :
F1=T-W1=T- M1*g
Net Force in M2
M2
M1
M1≠M2
F2=W2-T= M2*g-T
Apply Theory/Plan: In this question I should assume that there is no friction and the string has no mass that its tension
is the same throughout its length. Based on these assumptions, I know that unbalanced force acts on both body 1 and 2
because there is different magnitude of weight acting on each according to their different mass and there is no friction
that would cancel out the weight. Since the bodies are connected with the same string, I would also know that both
body will be affected by the same tension and share the same magnitude of velocity and acceleration. I know that
whatever net force acts, that Fnet = m*a, according to Newton’s Second Law, where I need to find the acceleration and
magnitude of tension. In order to find a I have to find mass and net force of the body. Fortunately I know that the mass
of the body is M1 and M2. For figuring out the net force for each body I have draw free body diagram for both object as
show above where I have identified that there is only gravity and tension working against each other vertically. From this
I can figure out the net force for each body equals to the difference between tension of the string and weight of each
object, then I will be able to produce the equation for each body that net force=ma. Then I would solve those two sets of
simultaneous equation of Newton’s second law by eliminating the unknown variables such as tension T, in order to get
the value of a in terms of g, M1 and M2. In a similar way I can solve for tension T, where instead of T I have to eliminate
variable acceleration a.
Formulas and Solve:
F1=T-W1
T- M1*g=M1*a
T=M1 g+M1 a
F2=W2-T
M2*g-T=M2*a
T=Mg-Ma
Fnet=ma
Solving for a
T1-T2= M1 g+M1 a -M2 g+M2 a
0=M1 g-M2 g+M1 a+M2 a
0=g(M1 -M2 )+a(M1 +M2 )
a(M1 +M2 )=g(M1 -M2 )
a=g(M1 -M2 )/( M1 +M2 )
15. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads
836 N. As the elevator moves up, the scale reading increases to 935 N. What is the acceleration of the
elevator during this time?
Diagram and Givens:
N
U=0
M=83.6
A=10
F=935
W
Apply Theory/Plan: I know that according to Newton’s second law of motion, when an unbalanced force acts on an
object the object will accelerate in the direction of the force according to the formula F=ma. Because of the reading on
the scale we know that the force is 935N. Also the scale initially reads 836N when the elevator is at rest which means
that is the mass. In order to find the mass we have to convert Newtons to Kilograms which we will do with the formula
w=ma. We can use -10 as the acceleration (due to gravity) because the elevator is at rest. Once we have the mass we
can finally solve for the acceleration with the formula from Newton’s second law of motion.
Formulas and Solve:
𝑊 =𝑚×𝑎
𝐹 =𝑚×𝑎
836 = 𝑚 × 10
83.6 = 𝑚
(935𝑁 = 83.6𝑘𝑔 × 𝑎) ÷ 10
(935 = 83.6𝑎) ÷ 10
𝑎 = (11.18) ÷ 10
𝑎 = 1.118
𝑎 = 1.12 m/s2
16. A 40 kg crate is pulled across the ice with a rope. A force of 100 N is applied at an angle of 30o with the
horizontal. Neglecting friction, calculate the acceleration of the crate and the upward force the ice exerts on
the crate as it is pulled.
Diagram and Givens:
F = 100 N
N
Fy
30°
Fx
m = 40 kg
F = 100 N
θ = 30°
m = 40 kg
W
Apply Theory/Plan: I know from Newton’s first law that objects will remain at rest or in constant velocity, therefore
have no change in motion, if there is no unbalanced force acting on them. Since an unbalanced force of 100 N is applied
to the crate, I know that there was a change in motion and therefore according to Newton’s Second Law, there must be
acceleration. I know that whatever net force acts, that Fnet = ma. Since I am given the force of 100 N applied at an angle
of 30° with the horizontal, I can find the net force acting on the crate, and from there plug the net force and the mass of
the crate into Fnet = ma, to find the acceleration that is needed to make this change in motion. The problem also asks to
find the upward force the ice exerts on the crate as it is pulled. I know that the equal and opposite forces referred to in
Newton’s third law of motion are acting on different objects. The force of the crate is acting on the ice, and the force of
the ice is acting on the crate. Since the crate is pulled across the ice with a rope at an angle, I know that the upward
force the ice exerts on the crate is less than the weight of the crate. I know that Fy + the normal force = weight.
Formulas and Solve:
Fcosθ = Fx
Fnet(x) = ma
F×sinθ = Fy
Fy + N = W
Acceleration:
100N×cos(30°) = 86.6N
86.6N = 40kg × a
86.8N/40kg = a
2.17m/s2 = a
2m/s2 ≈ a
Normal force:
100N×sin(30°) = 50N
50N + N = (40kg)(10m/s2)
2
= (40kg)(10m/s
) / 50N
Check: These answers make sense, because aNcrate
at 40kg weight
being pulled by a force of 100N at an angle would
= 350
Newtons
need quite an acceleration to change motion. TheNunits
work
out in each step, ending with m/s2 for the acceleration and
Newtons for the upward force the ice exerts on the crate as it is pulled.
17. Joe pushes the handle of a 10 kg fertilizer spreader. The handle makes a 45o with the horizontal. Joe wishes
to accelerate the spreader from rest to 1.39 m/s in 1.5 seconds. What force must Joe apply to the handle?
Diagram and Givens:
Normal
FApplied
45°
10 kg
Weight
U = 0 m/s, V = 1.39 m/s, t = 1.5 seconds, m = 10 kg
Apply theory/plan:
Newton’s first law states that an object in motion tends to stay in motion and an object at rest tends to stay at rest
unless acted upon by an external, unbalanced force. The question states that Joe wishes to accelerate the fertilizer
spreader and the mass is given, so we will be able to use Newton’s second law, which states that force is equal to mass
times acceleration. By looking at the diagram, we can also see that vectors will become useful, as the force Joe applies is
a vector with a given direction (45 degrees). Before beginning this problem, we can calculate the weight and normal
forces (which are equal, because the block is not moving in the vertical direction). This value has no specific significance
however, so calculating it is really unnecessary, unless we feel an urge to complete our free body diagram. We are given
the amount Joe wishes to accelerate the spreader and the mass of the spreader, so we can calculate the force needed to
accelerate the spreader in the x-direction fairly easily, but in the end we need to find the total force of the vector, so we
can use SOHCAHTOA to calculate the force Joe must apply.
Formulas and Solve:
𝑎=
𝑎=
𝑣−𝑢
𝑡
and F = ma
1.39𝑚/𝑠 − 0𝑚/𝑠 1.39𝑚/𝑠
=
= .92666𝑚/𝑠 2
1.5𝑠
1.5𝑠
Fx-direction = 10 kg * .92666 m/s2= 9.2666 kg m/s2 = 9.3 Newtons
Considering we need to find F, not Fx , we will have to define Fx in terms of F:
Cos(45) * F = Fx = 9.3
F = 13.2 = 13 Newtons
Check:
This answer makes sense as the force necessary is not enormous because the spreader only has a mass of 10 kg.
18. You slide a 325 N trunk up a 20.0o inclined plane with a constant velocity by exerting a force of 211 N
parallel to the inclined plane. What is the component of the trunk’s weight parallel to the incline? What is
the sum of your applied force, friction, and the parallel component of the trunk’s weight? What is the size
and direction of the frictional force?
Diagram and Givens:
Apply Theory/Plan: To find the answer to the first question I need to use vectors and trigonometry. I am told
to find the component of the trunk’s weight (Fw) parallel to the incline (Fwx). In other words, the x direction
component of the right angle shown with Fw. Next, I need to find the sum of the applied force, friction, and the
parallel component of the trunk’s weight. According to Newton’s first law, objects will remain at rest or in
uniform motion (constant velocity) provided no external unbalanced force acts on them. Since the trunk is
moving at constant velocity, and we know the force in one direction (the direction of the sliding), we know the
forces acting in the opposite direction must be balanced and equal. To find the size and direction of the
frictional force (FF), you must subtract the x component of the weight from the slide force. That will leave you
with the remaining force that is required for the trunk to be moving in uniform motion. The direction will be
parallel to the incline, and exactly opposite of the slide force
Formulas and Solve:
FX = 325 sin(20) = 111 N
211 N + (-211) = 0 N
211 N – 111 N = 100 N Parallel to incline, Opposite of the slide force. (180° to upward incline)