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.
Познато:
Сила (F) = 20 N
Поместување (s) = 2 m
Angle (θ) = 0
Wanted : Work (W)
Решение:
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!
Познат :
Сила (F) = 10 N
The horizontal force (Fx) = F cos 30o = (10)(0.5√3) = 5√3 N
Displacement (d) = 1 meter
Wanted : Work (W) ?
Решение :
W = Fx d = (5√3)(1) = 5√3 Joule
3. A body falls freely from rest, from a height of 2 m. If забрзување поради гравитацијата е 10 м/с2, determine the work done by the сила на гравитација!
Познато:
Object’s маса (м) = 1 кг
Висина (в) = 2 м
Забрзување поради гравитација (g) = 10 m/s2
Барани: Work done by the force of gravity (W)
Решение:
W = F d = wh = mgh
W = (1)(10)(2) = 20 Џули
W = work, F = force, d = distance, w = тежина, 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
Познато:
Маса (м) = 1 кг
Забрзување поради гравитација (g) = 10 m/s2
Elongation (x) = 2 cm = 0.02 m
Тежина (w) = mg = (1 kg) (10 m/s2) = 10 кг м/с2 = 10 Н.
Барани: Spring constant and work done by spring force
Решение:
(a) Spring constant
Формула на Закон на Хук :
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
W = – ½ kx2
W = – ½ (500)(0.02)2
W = – (250)(0.0004)
W = -0.1 Џул
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 сила на триење Fk = 2 N. Determine the net work done on the box.
Познато:
Сила (F) = 10 N
Force of kinetic friction (Fk) = 2 N
Displacement (d) = 2 m
Барани: Net work (Wнето)
Решение:
Work done by force F :
W1 = F d cos 0 = (10)(2)(1) = 20 Joule
Work done by force of kinetic friction (Fk):
W2 = Ф.k d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule
Net work :
Wнето = В1 - В2
Wнето = 20 - 4
Wнето = 16 џули
6. What is the work done by force F on the block.
Познато:
Сила (F) = 12 Њутн
Displacement (d) = 4 meters
Сакаше: Работа (W)
Решение:
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?
Познато:
Сила (F) = 200 Њутн
Displacement (d) = 2 meters
Сакаше: Работа (W)
Решение:
Работа:
W = F s
W = (200 Newton)(2 meters)
W = 400 N m
W = 400 џули
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?
Познато:
Displacement (d) = 10 meters – 0.5 meters = 9.5 meters
Сила (F) = 50 Њутн
Барани: Работа (W)
Решение:
W = F s
W = (50 Newton)(9.5 meters)
W = 475 N m
W = 475 џули
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.
Познато:
Displacement (s) = 4 meters
Net force (F) = 50 Newton + 70 Newton = 120 Newton
Сакаше: Работа (W)
Решение:
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?
Познато:
Сила (F) = 250 Њутн
Displacement (s) = 1000 cm = 1000/100 meters = 10 meters
Барани: Работа (W)
Решение:
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.
Познато:
работа (W) = 375 Joule
јачина на нето (ΣF) = 40 N + 10 N – 25 N = 25 Newton (десно)
Барани: Поместување (d)
Решение:
The equation of work :
W = F s
Object’s displacement :
d = W / F = 375 Joule / 25 Newton
d = 15 метриs
12. The activities below wкој не прави работи 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
Решение:
The equation of work :
W = ΣF s
W= работи, F = сила, d = поместување
Based on the above formula, work done by force and there is a displacement.
Точниот одговор е 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 кружни движења.
А. 0 џули
Б. 1400 џули
C. 1540 џули
D. 1760 џули
Решение:
If the person pushes инвалидска количка 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.
Точен одговор е А.
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.
А. 45 J
Б. 72 Ј
C. 1680 J
Д. 2580 J
Познато:
The force of push (F) = 350 Newton
Сила на триење (Fфрик) = 70 Њутн
Displacement of object (s) = 6 meters
Сакаше: Работа (W)
Решение:
There are two forces that act on the object, the push force (F) and friction force (Fфрик). 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 = – (Fфрик)(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule
The net work :
W net = 2100 Joule – 420 Joule
W net = 1680 Joule
Точниот одговор е 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.
А. 0.5 џули
Б. 3 џули
C. 32 џули
D. 192 џули
Познато:
Push force (F) = 14 Newton
Сила на триење (Fфрик) = 10 Њутн
Displacement of object (d) = 8 meters
Сакаше: Работа (W)
Решение:
There are two forces that act on an object, push force (F) and friction force (Fфрик).
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 = – (Fфрик)(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule
The net work :
W net = 112 Joule – 80 Joule
W net = 32 Joule
Точниот одговор е C.
16. Determine the net work based on figure below.
А. 360 џули
Б. 450 џули
C. 600 џули
D. 750 џули
Решение:
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
Точен одговор е А.
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
Б. 16
В 28
D. 30
Познато:
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
Барани: The wood will enter entirely into the ground after…. hits.
Решение:
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
Точниот одговор е Д.
[wpdm_package id='1192']
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