Fisikastudycenter.com - Sample Questions and Discussion on Motion
Dynamics, Condensed Matter Physics 2 class (XI) High School involving
friction dynamics in some cases, variations in determining the value of
the normal force, contact force and the decomposition of styles. Formulas - Formula Minimal
Newton's Law I
Σ F = 0
→ stationary objects or
→ object moving at a constant speed / fixed or
→ zero acceleration of the motion of objects or
→ object moving uniformly straight (GLB)
Newton's second law
Σ F = ma
→ object moving with constant acceleration
→ object moving uniformly accelerated (GLBB)
→ velocity of objects change
Swipe style
Static Friction Style → f s = μ s N
Kinetic Friction Style → f k = μ k N
with N = normal force, μ s = coefficient of static friction, μ k = coefficient of kinetic friction
Gravity
W = mg
Example Problem and Discussion
No Problem. 1
Consider the following picture!
Body of mass m = 10 kg is pulled over rough floor by a force F = 12 N to the right. If the coefficient of static friction between the object and the floor is 0.2 with a coefficient of kinetic friction 0.1 magnitude specify:
a) normal force
b) The force of friction between the object and the floor
c) Acceleration of the motion of objects
Discussion
The forces on the object shown in the following figure:
a) normal force
Σ F y = 0
N - W = 0
N - mg = 0
N - (10) (10) = 0
N = 100 N
b) The force of friction between the object and the floor
Check beforehand that the maximum static friction can occur between the object and the floor:
f s max = μ s N
f s max = (0.2) (100) = 20 N
It turns out the maximum static friction is greater than the force that pulls objects (F) so that the object is at rest. In accordance with Newton's law for stationary objects:
Σ F x = 0
F - f ges = 0
12 - ges f = 0
ges f = 12 N
c) Acceleration of the motion of objects
Idle objects, object acceleration NOL
No Problem. 2
Consider the following figure, first thing in the break condition!
Body of mass m = 10 kg is pulled over rough floor by a force F = 25 N to the right. If the coefficient of static friction between the object and the floor is 0.2 with a coefficient of kinetic friction 0.1 magnitude specify:
a) normal force
b) The force of friction between the object and the floor
c) Acceleration of the motion of objects
d) the distance of objects after 2 second
Discussion
The forces on the object shown in the following figure:
a) normal force
Σ F y = 0
N - W = 0
N - mg = 0
N - (10) (10) = 0
N = 100 N
b) The force of friction between the object and the floor
Check beforehand that the maximum static friction can occur between the object and the floor:
f s max = μ s N
f s max = (0.2) (100) = 20 N
It turns out the maximum static friction force (20 N) is smaller than the force that pulls objects (25 N), so that moving objects. For moving objects geseknya force is friction with coefficient of kinetic friction:
ges f = f k = μ k N
ges f = (0.1) (100) = 10 N
c) Acceleration of the motion of objects
Newton's second law:
Σ F x = ma
F - f = ma ges
25-10 = 10a
a = 15/10 = 1.5 m / s 2
d) the distance of objects after 2 second
S = V o t + 1/2 at 2
S = 0 + 1/2 (1.5) (2 2)
S = 3 meters
No Problem. 3
Consider the following figure, 5 kg object initially in a state of not moving!
If the angle formed between the force F = 25 N with a horizontal line is 37 °, the coefficient of kinetic friction is 0.1 and the floor surface gravity acceleration of 10 m / s 2 specify the value:
a) normal force
b) Friction
c) Acceleration of the motion of objects
(Sin 37 o = 0.6 and cos 37 o = 0.8)
Discussion
The forces on the object shown in the following figure:
a) normal force
Σ F y = 0
N + F sin θ - W = 0
N = W - F sin θ = (5) (10) - (25) (0.6) = 35 N
b) Friction
If the problem is only known to the kinetic friction coefficient, then certainly object to move, so that f = f k ges:
ges f = μ k N
ges f = (0.1) (35) = 3.5 N
c) Acceleration of the motion of objects
Σ F x = ma
F cos θ - f = ma ges
(25) (0.8) - 3.5 = 5a
5a = 16.5
a = 3.3 m / s 2
No Problem. 4
Consider the following figure, 100 kg beam launched from a hill!
Think of the hillside and the average coefficient of friction of 0.125. Earth's gravitational acceleration of 10 m / s 2 and sin 53 o = 0.8, cos 53 o = 0.6. Determine the value of:
a) the normal force on the beam
b) Friction between the slope and the beam
c) Acceleration of the beam motion
Discussion
The forces on the beam are shown in the following figure:
a) the normal force on the beam
Σ F y = 0
N - W cos θ = 0
N - mg cos 53 o = 0
N - (100) (10) (0.6) = 0
N = 600 Newton
b) Friction between the slope and the beam
ges f = μ k N
ges f = (0.125) (600) = 75 newtons
c) Acceleration of the beam motion
Σ F x = ma
W sin θ - f = ma ges
mg sin 53 o - f = ma ges
(100) (10) (0.8) - 75 = 100a
a = 725/100 = 7.25 m / s 2
No Problem. 5
A beam of mass 40 kg and block B of mass 20 kg is above the slippery surface driven by a force F of 120 N as shown in the following picture!
Define:
a) Acceleration of motion both beams
b) contact force occurring between beams A and B
Discussion
a) Acceleration of motion both beams
Review the system:
Σ F = ma
120 = (40 + 20) a
a = 120/60 m / s 2
b) contact force occurring between beams A and B
The first way, Review of object A:
Σ F = ma
F - F = m A a contact
120 - F contacts = 40 (2)
Contact F = 120-80 = 40 Newton
The second way, Review of object B:
Σ F = ma
F = m B a contact
Contact F = 20 (2) = 40 Newton
No Problem. 6
Beams A and B lie on the surface slippery incline driven by a force F of 480 N as shown in the following picture!
Define:
a) Acceleration of motion both beams
b) contact force between the beams A and B
Discussion
a) Acceleration of motion both beams
Review System:
The forces on the two objects (A and B combined) shown in the following figure:
Σ F = ma
F - W sin 37 o = ma
480 - (40 + 20) (10) (0.6) = (40 + 20) a
a = 120/60 = 2 m / s 2
b) contact force between the beams A and B
The first way, review the beams A
The forces on the beam A is shown in the following figure:
Σ F = ma
F - W A sin 37 o - F = m A a contact
480 - (40) (10) (0.6) - F contacts = (40) (2)
480-240 - 80 = F contacts
F = 160 Newton contacts
The second way, review the objects B
Σ F = ma
F contacts - W B sin 37 o = m B a
F contact - (20) (10) (0.6) = (20) (2)
Contact F = 40 + 120 = 160 Newton
No Problem. 7
A beam weighing 100 N with a rope tied horizontally in C (see figure). Block B weighs 500 N. Coefficient of friction between A and B = 0.2 and the coefficient of friction between B and the floor = 0.5. Minimum magnitude of the force F to move the beam B is .... newton
A. 950
B. 750
C. 600
D. 320
E. 100
(Source Problem: UMPTN 1993)
Discussion
AB → f friction between block A and B
BL → f friction between block B and the floor
f AB AB N = μ
f AB = (0.2) (100) = 20 N
BL BL f = μ N
f BL = (0.5) (100 + 500) = 300 N
Review the object B
Σ F x = 0
F - f AB - f BL = 0
F - 20-300 = 0
F = 320 Newton
No Problem. 8
The first object with mass m1 = 6 kg and a second object with mass m2 = 4 kg are connected by a pulley slick look at the picture below!
If the floor is slippery and m 2 is pulled to the right force F = 42 Newton, specify:
a) Acceleration of the first object
b) Acceleration of the second object
c) Voltage strap T
Discussion
a) Acceleration of the first object
Relationship between the acceleration of the object first (a 1) and the acceleration of both objects (a 2) are:
a 1 = 2a 2
or
a 2 = 1/2 a 1
Review m 2
F - 2T = m 2 a 2
42 - 2T = 4a 2
42 - 2T = 4 (1/2) a 1
42 - 2T = 2a 1 (Pers. 1)
Review the m 1
T = m 1 a 1
T = 6 A 1 (Pers. 2)
Join the Press. 1 and Eq. 2
42 - 2T = 2a 1
42-2 (6a 1) = 2a 1
42 = 14, a 1
a 1 = 42/14 = 3 m / s 2
b) Acceleration of the second object
a 2 = 1/2 a 1
a 2 = 1/2 (3) = 1.5 m / s 2
c) Voltage strap T
T = 6a 1 = 6 (3) = 18 Newton
No Problem. 9
A mass = 4 kg, B = 6 kg mass is connected by a rope and pulled a force F = 40 N to the right at an angle of 37 ° to the horizontal direction!
If the coefficient of kinetic friction with the floor of the two masses is 0.1 specify:
a) Acceleration of mass motion of the two
b) The voltage between the rope connecting the two masses
Discussion
B mass review:
Value of the normal force N:
Σ F y = 0
N + F sin 37 o = W
N + (40) (0.6) = (6) (10)
N = 60-24 = 36 N
Large friction:
ges f B = μ k N
ges f B = (0.1) (36) = 3.6 N
Newton's second law:
Σ F x = ma
F cos 37 o - f ges B - T = ma
(40) (0.8) - 3.6 - T = 6 a
28.4 - T = 6 a → (equation 1)
Overview of the forces on the mass A
Σ F x = ma
T - A ges f = ma
T - μ k N = ma
T - μ k mg = ma
T - (0.1) (4) (10) = 4 a
T = 4a + 4 → Equation 2
Join 1 and 2
28.4 - T = 6 a
28.4 - (4a + 4) = 6 a
24.4 = 10a
a = 2.44 m / s 2
b) The voltage between the rope connecting the two masses
T = 4a + 4
T = 4 (2.44) + 4
T = 13.76 Newton
No Problem. 10
Given the following picture!
If the mass of the pulley is negligible, determine:
a) Acceleration of the second movement of objects
b) Voltage rope connecting the two objects
Discussion
A Preview
Σ F x = ma
T - W A sin 37 o = m A a
T - (5) (10) (0.6) = 5 a
T - 30 = 5a → (Equation 1)
Review B
Σ F x = ma
W B sin 53 o - T = m B a
(10) (0.8) - T = 10 a
T = 80-10 a → (Equation 2)
Join 1 and 2
T - 30 = 5a
(80-10 A) - 30 = 5 a
15 a = 50
a = 50/15 = 10/3 m / s 2
b) Voltage rope connecting the two objects
T - 30 = 5a
T - 30 = 5 (10/3)
T = 46.67 Newton
No Problem. 11
Image given as follows:
A block of mass = 6 kg, the mass of block B = 4 kg. Coefficient of kinetic friction between block A and B is 0.1 and the coefficient of friction between block A and the floor is 0.2. Determine the magnitude of force F so that A moves straight uniform beam to the right, ignore the mass of the pulley!
Discussion
Review B
Objects moving uniformly straight → a = 0
Σ F x = 0
T - f BA = 0
T = f BA = μ μ BA BA N = mg = (0.1) (4) (10) = 4 N
A Preview
Σ F x = 0
F - T - F AB - f AL = 0
with f = μ AL AL N = (0,2) (10) (10) = 20 N
(Normal force on A is the weight of the total weight of A plus B, because stacked)
So that:
F - 4 - 4-20 = 0
F = 28 Newton
No Problem. 12
An elevator of mass 400 kg moves vertically upward from rest with constant acceleration of 2 m / s 2. If the acceleration of gravity is 9.8 m / s 2, then the towing rope tension elevators is ....
A. 400 newton
B. 800 newton
C. 3120 newton
D. 3920 newton
E. 4720 newton
(Source Problem: I Pilot Project 1981)
Discussion
Σ F y = ma
T - W = ma
T - (400) (9.8) = (400) (2)
T = 800 + 3920 = 4720 Newton
No Problem. 13
Of about 12 numbers, specify voltage towing rope elevator if elevator movement is down!
Discussion
Elevators move down:
Σ F y = ma
W - T = ma
(400) (9.8) - T = (400) (2)
T = 3920 - 800 = 3120 Newton
No Problem. 14
Note the arrangement of the following two items:
Coefficient of kinetic friction between the mass of the first floor is 0.1, the mass of the first object = 4 kg and 6 kg mass of the two objects. Define:
a) Acceleration of the motion of objects first
b) Acceleration of the motion of objects both
Discussion
a) Acceleration of the motion of objects first
Acceleration of the object relationship first and second objects are:
a 1 = 2a 2
or
a 2 = 1/2 a 1
Review the first object
Σ F x = m 1 a 1
T - f = 4 a 1
T - μ k N = 4a 1
T - (0.1) (4) (10) = 4 a 1
T = 4a 1 + 4 → Equation 1
Review the second object
Σ F y = m 2 a 2
W - 2T = (6) (1/2 a 1)
60 - 2T = 3a 1 → Equation 2
Join Equation 2 and Equation 1
60 - 2T = 3 a 1
60-2 (4a 1 + 4) = 3a 1
60 - 8a 1-8 = 3a 1
52 = 11a 1
a 1 = 52/11 m / s 2
b) Acceleration of the motion of objects both
a 2 = 1/2 a 1
a 2 = 1/2 (52/11) = 26/11 m / s 2
No Problem. 15
M beam of mass 10 kg attached to the rough walls with kinetic friction coefficient of 0.1. Beam gets horizontal force F 2 = 50 N and a vertical force F 1.
Determine the magnitude of the vertical force F 1 so that the beam moves vertically upward with an acceleration of 2 m / s 2!
Discussion
Overview of the forces acting on m:
Σ F x = 0
N - F 2 = 0
N - 50 = 0
N = 50 Newton
Σ F y = ma
F 1 - W - f = ma
F 1 - mg - μ k N = ma
F 1 - (10) (10) - (0.1) (50) = 10 (2)
F 1 = 20 + 100 + 5 = 125 Newton
Problem Solved
Solved Problem # 1
Body of mass 4 kg given initial velocity of 10 m / s from the lower end of the inclined plane as shown.
Objects experience friction from the field at 16 N and sinα = 0.85. Objects stop after a distance
(A) 3 m
(B) 4 m
(C) 5 m
(D) 6 m
(E) 8 m
(Source Problem: UM UGM 2009)
No matter Exercise. 2
Five of the object (call it blocks), each of mass 2 kg, 3 kg, 4 kg, 5 kg and 6 kg, connected by ropes without mass (fine), and then pulled flat on the floor with a force of 40 N as the picture below.
Coefficient of friction between each object and the floor 0.1, the gravitational acceleration of 10 m / s 2. Determine the magnitude of the voltage cord connecting objects:
a) 2 kg and 3 kg
b) 4 kg and 5 kg
(Image source: UM UGM 2008)
