1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block.
Poznato:
Sila (F) = 20 N
premještanje (s) = 2 m
Ugao (θ) = 0
Wanted : Work (W)
Rešenje:
W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule
2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30o angle as shown in figure below. Determine the work done by force F!
Poznat :
Sila (F) = 10 N
The horizontal force (Fx) = F cos 30o = (10)(0.5√3) = 5√3 N
Displacement (d) = 1 meter
Wanted : Work (W) ?
rastvor :
Š = Žx d = (5√3)(1) = 5√3 Joule
3. A body falls freely from rest, from a height of 2 m. If ubrzanje zbog gravitacije je 10 m/s2, determine the work done by the sila gravitacije!
Poznato:
Objekta Masa (m) = 1 kg
Visina (v) = 2 m
Ubrzanje usljed gravitacije (g) = 10 m/s2
Traži se: Work done by the force of gravity (W)
Rešenje:
W = F d = w h = m g h
W = (1)(10)(2) = 20 Džula
W = work, F = force, d = distance, w = težina, h = height, m = mass, g = acceleration due to gravity.
4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m/s2, determine (a) the spring constant (b) work done by spring force on object
Poznato:
Masa (m) = 1 kg
Ubrzanje usljed gravitacije (g) = 10 m/s2
Izduženje (x) = 2 cm = 0.02 m
Težina (w) = mg = (1 kg)(10 m/s2) = 10 kg m/s2 = 10 N
Traži se: Spring constant and work done by spring force
Rešenje:
(a) Spring constant
Formula od Hookeov zakon :
F = kx.
k = F / x = w / x = mg / x
k = (1)(10) / 0.02 = 10 / 0.02
k = 500 N/m
(b) work done by spring force
Š = – ½ kx2
W = – ½ (500)(0.02)2
W = – (250)(0.0004)
W = -0.1 Džul
The minus sign indicates that the direction of spring force is opposite with the direction of object displacement.
5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a sila trenja Fk = 2 N. Determine the net work done on the box.
Poznato:
Sila (F) = 10 N
Force of kinetic friction (Fk) = 2 N
Displacement (d) = 2 m
Traži se: Net work (Wneto)
Rešenje:
Work done by force F :
W1 = F d cos 0 = (10)(2)(1) = 20 Joule
Work done by force of kinetic friction (Fk):
W2 =Fk d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule
Net work :
Wneto =W1 - W2
Wneto = 20 - 4
Wneto = 16 džula
6. What is the work done by force F on the block.
Poznato:
Sila (F) = 12 Newtona
Displacement (d) = 4 meters
Wanted: Rad (W)
Rešenje:
W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule
7. A block is pushed by a force of 200 N. The block’s displacement is 2 meters. What is the work done on the block?
Poznato:
Sila (F) = 200 Newtona
Displacement (d) = 2 meters
Wanted: Rad (W)
Rešenje:
posao :
W = F s
W = (200 Newton)(2 meters)
W = 400 N m
W = 400 Džula
8. The driver of the sedan wants to park his car exactly 0.5 m in front of the truck which is at 10 m from the sedan’s position. What is the work required by the sedan?
Poznato:
Displacement (d) = 10 meters – 0.5 meters = 9.5 meters
Sila (F) = 50 Newtona
Traži se: Rad (W)
Rešenje:
W = F s
W = (50 Newton)(9.5 meters)
W = 475 N m
W = 475 Džula
9.
Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N.
Poznato:
Displacement (s) = 4 meters
Net force (F) = 50 Newton + 70 Newton = 120 Newton
Wanted: Rad (W)
Rešenje:
W = F s = (120 Newton)(4 meters) = 480 N m = 480 Joule
10. A driver pulling a car so the car moves as far as 1000 cm. What is the work done on the car?
Poznato:
Sila (F) = 250 Newtona
Displacement (s) = 1000 cm = 1000/100 meters = 10 meters
Traži se: Rad (W)
Rešenje:
W = F s = (250 Newton)(10 meters) = 2500 N m = 2500 Joule
11. Based on figure below, if work done by net force is 375 Joule, determine object’s displacement.
Poznato:
rad (W) = 375 Joule
Neto sila (ΣF) = 40 N + 10 N – 25 N = 25 Newton (desno)
Traži se: premještanje (d)
Rešenje:
The equation of work :
W = F s
Object’s displacement :
d = W / F = 375 Joule / 25 Newton
d = 15 metaras
12. The activities below which ne radi rad is ...
A. Push an object as far as 10 meters
B. Push a car until a move
C. Push a wall
D. Pulled a box
Rešenje:
The equation of work :
W = ΣF s
W= rad, F = sila, d = raseljavanje
Based on the above formula, work done by force and there is a displacement.
Tačan odgovor je C.
13. Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times kružno kretanje.
A. 0 Džula
B. 1400 džula
C. 1540 Džula
D. 1760 Džula
Rešenje:
If the person pushes invalidska kolica for two times circular motion then the person and wheelchair return to the original position, so the displacement of the person is zero.
Displacement = 0 so work = 0.
Tačan odgovor je A.
14. Someone push an object on the floor with force of 350 N. The floor exerts a friction force 70 N. Determine the work done by force to move the object as far as 6 meters.
A. 45 J
B. 72 J
Oko 1680 J
D. 2580 J
Poznato:
The force of push (F) = 350 Newton
Sila trenja (Ffric) = 70 Njutna
Displacement of object (s) = 6 meters
Wanted: Rad (W)
Rešenje:
There are two forces that act on the object, the push force (F) and friction force (Ffric). The push force has the same direction as the displacement of the object because the push force does a positive work. In another hand, the friction force has the opposite direction with a displacement of the object so that the friction force does a negative work.
Work done by push force :
W = F d = (350 Newton)(6 meters) = 2100 Newton-meters = 2100 Joule
Work done by friction force :
W = – (Ffric)(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule
The net work :
W net = 2100 Joule – 420 Joule
W net = 1680 Joule
Tačan odgovor je C.
15. An object is pushed by a horizontal force of 14 Newton on a rough floor with the friction force of 10 Newton. Determine the net work of move the object as far as 8 meters.
A. 0.5 Džula
B. 3 džula
C. 32 Džula
D. 192 Džula
Poznato:
Push force (F) = 14 Newton
Sila trenja (Ffric) = 10 Njutna
Displacement of object (d) = 8 meters
Wanted: Rad (W)
Rešenje:
There are two forces that act on an object, push force (F) and friction force (Ffric).
The push force has the same direction as the displacement of the object so that the push force does a positive work. In another hand, the friction force has the opposite direction as the displacement of the object so that the friction force does a negative work.
Work done by push force :
W = F s = (14 Newton)(8 meters) = 112 Newton meters = 112 Joule
Work done by friction force :
W = – (Ffric)(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule
The net work :
W net = 112 Joule – 80 Joule
W net = 32 Joule
Tačan odgovor je C.
16. Determine the net work based on figure below.
A. 360 Džula
B. 450 džula
C. 600 Džula
D. 750 Džula
Rešenje:
Work = Force (F) x displacement (d)
Work = Area of triangle 1 + area of rectangle + area of triangle 2
Work = 1/2(40-0)(3-0) + (40-0)(9-3) + 1/2(40-0)(12-9)
Work = 1/2(40)(3) + (40)(6) + 1/2(40)(3)
Work = (20)(3) + 240 + (20)(3)
Work = 60 + 240 + 60
Work = 360 Joule
Tačan odgovor je A.
17. A piece of wood with a length of 60 cm plugged vertically into the ground. Wood hit with a 10-kg hammer from a height of 40 cm above the top of the wood. If the average resistance force of the ground is 2 x 103 N and the acceleration due to gravity is 10 m/s2, then the wood will enter entirely into the ground after…. hits.
A. 4
B. 16
C. 28
D. 30
Poznato:
Mass of hammer (m) = 10 kg
Acceleration due to gravity (g) = 10 m/s2
Weight of hammer (w) = m g = (10)(10) = 100 kg m/s2
Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters
The resistance of wood (F) = 2 x 103 N = 2000 N
Length of wood (s) = 60 cm = 0.6 meters
Traži se: The wood will enter entirely into the ground after…. hits.
Rešenje:
Work done on the hammer when hammer moves as far as 0.4 meters is :
W = F d = w s = (100 N)(0.4 m) = 40 Nm = 40 Joule
Work done by the resistance force of the ground :
W = F d = (2000 N)(0.6 m) = 1200 Nm = 1200 Joule
The wood will enter entirely into the ground after…. hits.
1200 Joule / 40 Joule = 30
Tačan odgovor je D.
[wpdm_package id='1192′]
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