1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetičko trenje je 0.2. Ubrzanje zbog gravitacije je 10 m/s2. What is the magnitude of raseljavanje?
Known ;
masa (m) = 2 kg
Početna brzina (vo) = 10 m/s
Konačna brzina (vt) = 0 m/s
The coefficient of kinetic friction (μk) = 0.2
Težina (w) = mg = (1 kg)(10 m/s2) = 10 kg m/s2 = 10 N
Traži se: The magnitude of displacement (d)
Rešenje:
Princip rada i mehaničke energije :
Wnc = ΔEM
Wnc = ΔEK + ΔEP
Wnc = ½ m (vt2 - vo2) + m g h
Wnc = rad uradio nonconservative force acting on the object
ΔEK = the change in kinetička energija
ΔEP = the change in potencijalna energija
m = masa
v = brzina
g = acceleration due to gravity
h = the change in height
The change in height h = 0 so ΔEP = 0
Wnc = ΔEK
Work done by nonconservative force :
Wnc = -Fc d = – μk N d = – μk w d = – μk m g d
Wnc = -(0.2)(2)(10)(s)
Wnc = – (4)(2)
The minus sign indicates that the direction of kinetic friction force is opposite with the direction of displacement.
Promjena kinetičke energije:
ΔEK = ½ m (vt2 - vo2) = ½ (2)(0 – 102) = 0 – 100 = -100
The magnitude of displacement :
Wnc = ΔEK
– 4 s = – 100
s = -100/-4
s = 25 m
2. A block slides down on inclined plane. The coefficient of kinetic friction is 0.4. Acceleration due to gravity is g = 10 ms-2. Determine the final velocity when the block hits the ground.
Poznato:
Početna visina (ho) = 6 m
Konačna visina (ht) = 0 m
Početna brzina (vo) = 0
The coefficient of kinetic friction (μk) = 0.4
Ubrzanje usljed gravitacije (g) = 10 ms-2
cos θ = 8 / 10
The vertical component of weight = wy = w cos θ = m g cos θ = m (10)(8/10) = m (10)(4/5) = m (40/5) = 8 m
Normal force = N = wy = 8 m
Force of kinetic friction = fk = μk N= μk wy = (0.4)(8 m) = 3.2 m
Traži se: Konačna brzina (vt)
Rešenje:
The work-mechanical energy principle :
Wnc = ΔEM
Wnc = ΔEK + ΔEP
Promjena kinetičke energije:
ΔEK = 1/2 m (vt2 - vo2) = 1/2 m (vt2 - 0) = 1/2 mVt2
Promjena potencijalne energije:
ΔEP = mg (ht - ho) = m(10)(0-6) = m(10)(-6) = –60 m
Work done by force of kinetic friction :
Wnc = – fk s = – (3.2 m)(10) = – 32 m
The minus sign indicates that the direction of force of kinetic friction is opposite with the direction of displacement.
Konačna brzina (vt):
Wnc = ΔEK + ΔEP
- 32 m = 1/2 mVt2 - 60 m
- 32 m = m (1/2 vt2 - 60)
- 32 = 1/2 vt2 - 60
– 32 + 60 = 1/2 vt2
28 = 1/2 vt2
2 (28) = vt2
56 = vt2
vt = √4.14
vt = 2√14 ms-1
3. A block slides down on rough inclined plane. The initial velocity is 0 m/s and the final velocity is 10 ms-1. If the force of kinetic friction is 2 N and acceleration due to gravity g = 10 ms-2, what is height (h) ?
Poznato:
Masa (m) = 1 kg
Početna brzina (vo) = 0 (block rest)
Konačna brzina (vt) = 10 ms-1
Početna visina (ho) = h
Konačna visina (ht) = 0
Sila kinetičkog trenja (fk) = 2 N
Ubrzanje usljed gravitacije (g) = 10 ms-2
Traži se: visina (h)
Rešenje:
Work done by the force of kinetic friction :
Wnc = – fk s = – (2)(15) = – 30
The minus sign indicates that the direction of force of kinetic friction is opposite with the direction of displacement.
Promjena kinetičke energije:
ΔEK = 1/2 m (vt2 - vo2) = 1/2 (1)(102 - 0) = 1/2 (102) = 1/2 (100) = 50
Promjena potencijalne energije:
ΔEP = mg (ht - ho) = (1)(10)(0-h) = (10)(-h) = -10 h
The work-mechanical energy principle :
Wnc = ΔEK + ΔEP
– 30 = 50 – 10 sati
10 sati = 50 + 30
10 h = 80
h = 80/10
h = 8m
4. If a block moves down a roughly inclined plane, then ….
A. The work done by gravity force on the block is greater than the change of potential energy of the block
B. The mechanical energy increases
C. The amount of kinetic energy and its potential energy is reduced
D. The work done by the friction force equal to the change in kinetic energy of the block
rastvor
A is wrong
Work done by the force of gravity on the block equal to the change in the gravitational potential energy of a block.
B is wrong
Ako je inclined plane is smooth then the mechanical energy of the block is constant. When at the top of the inclined plane and not yet moves, the mechanical energy of the block is equal to the gravitational potential energy. The block still at rest so its kinetic energy is zero. When moves down an inclined plane, the height of the block is reduced so the gravitational potential energy is also reduced. The gravitational potential energy decreases because it changed to the kinetic energy. Although the gravitational potential energy change into the kinetic energy but the mechanical energy is constant.
If the inclined plane is rough then the mechanical energy of the block is decreased because of the negative work done by the friction force. The friction force is a non-conservative force. Work done by a non-conservative force on an object causes the mechanical energy of the object is decreased.
C is correct
The inclined plane is rough so there is a friction force that challenges the motion of the block, The friction force is a non-conservative force. Theorem work-mechanical energy states that work done by a non-conservative force (for example friction force) equal to the change of the mechanical energy. In this chase, the mechanical energy is decreased.
The mechanical energy of block = the gravitational potential energy + the kinetic energy.
D is wrong.
The work done by the friction force on the block equal to the change of the mechanical energy of block, not the change of the kinetic energy of the block. True that the friction force challenges the motion of the block so it decreases the speed of block and decreases the kinetic energy of the block. But realize that the kinetic energy of the block comes from the gravitational potential energy. So it’s true stated that the work done by the friction force equal to the change in the mechanical energy (mechanical energy decreases).
Tačan odgovor je C.
5. An object on a rough floor is hit so it moves for 3 seconds then stop. If the known mass of the object is 10 grams, the friction force between object and floor is 2 kilodyne. Determine the work done by the friction force.
A. 0.18 J
B. -0.18 J
Oko 0.36 J
D. -0.36 J
Poznato:
Vremenski interval (t) = 3 sekundi
Konačna brzina (vt) = 0 m/s (object rests)
Mass of object (m) = 10 grams = 10/1000 kg = 1/100 kg = 0.01 kg
Friction force (F) = 2 kilodyne = 2 x 103 dyne
Traži se: Work (W) done by friction force
Rešenje:
Conversion of unit of force :
1 Newton = 1 x 105 dyne
1 dyne = 1 / 105 Newton = 1 x 10-5 Newton = 10-5 njutn
Friction force (F) = 2 x 103 dyne = 2 x 103 x 10-5 Newton = 2 x 10-2 Newton = 2/100 Newton = 0.02 Newton
Theorem work-mechanical energy states that work done by a non-conservative force on an object equal to the change in the mechanical energy of the object.
If an object moves on the inclined plane then the mechanical energy (ME) = the gravitational potential energy (PE) + the kinetic energy (KE). But if the object moves just on a horizontal plane so there is no change in height the mechanical energy = the kinetic energy. On the horizontal plane, the gravitational potential energy is zero because there is no change in height.
The friction force is a non-conservative force. The friction force usually decreases the object’s speed and an object’s kinetic energy. Can conclude that work done by the friction force on an object equal to the decreases of the mechanical energy.
Mathematically :
Work (W) = The change of the mechanical energy (ΔEM)
F s = ΔEP + ΔEK
F s = m g Δh + ½ m (vt2 - vo2)
F s = m g (0) + ½ m (02 - vo2)
F s = 0 + ½ m (– (vo2))
F s = – ½ m vo2 ———— Equation 1
Description : F = friction force, d = displacement, m = mass, vo = početna brzina
Displacement (d) and initial speed (vo) not known yet because first calculate vo or d. Work done by the friction force calculated using one of the equation after known vo or d.
The equation of displacement (d) on nonuniform linear motion :
vt2 = vo2 + 2 (-a) s
Konačna brzina (vt) = 0 and acceleration (a) is signed negative because the object is decelerated (the speed of object is decreases).
0 = vo2 – 2 a d
vo2 = 2 a d
d = vo2 / 2 a ———— Equation 2
Change d on equation 2 with d in equation 1 :
F d = – ½ m vo2
F (vo2/2a) = – ½ m vo2
(F/2a) vo2 = – ½ m vo2
F/2a = – ½ m
F = (2)(a)(-1/2)(m)
F = – (a)(m)
a = – (F / m)
a = – 0.02 Newton / 0.01 kilogram
a = – 2 Newton/kilogram
a = – 2 m/s2
Equation to calculate the initial speed (vo) in nonuniform linear motion :
vt = vo + a t —–> Final speed (vt) = 0, Acceleration (a) = 2 m/s2, time interval (t) = 3 seconds
0 = vo + (-2)(3)
0 = vo - 6
vo = 6 meters/second
Work (W) done by the friction force :
W = – ½ m vo2 = -1/2 (0.01)(62) = -1/2 (0.01)(36)
W = -1/2 (0.36)
W = – 0.18 Joule
Work done by the friction force is signed negative means that the work decreases the mechanical energy of the object.
If known d the work can calculate using the equation of W = F d.
Tačan odgovor je B.
[wpdm_package id='1178′]
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- Problemi i rješenja principa rada i mehaničke energije
- Problemi i rješenja gravitacione potencijalne energije
- Problemi i rješenja potencijalne energije elastičnih opruga
- Problemi s napajanjem i rješenja
- Primjena zakona o očuvanju mehaničke energije za kretanje slobodnog pada
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