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.
Cunnisciutu:
Forza (F) = 20 N
Sustantimentu (s) = 2 m
Angle (θ) = 0
Wanted : Work (W)
Soluzione:
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!
Cunnisciutu :
Forza (F) = 10 N
The horizontal force (Fx) = F cos 30o = (10)(0.5√3) = 5√3 N
Displacement (d) = 1 meter
Wanted : Work (W) ?
Vergogna à tè :
W = Fx d = (5√3)(1) = 5√3 Joule
3. A body falls freely from rest, from a height of 2 m. If accelerazione per via di a gravità hè 10 m/s2, determine the work done by the forza di gravità!
Cunnisciutu:
Object’s di massa (m) = 1 kg
Altezza (h) = 2 m
Accelerazione di gravità (g) = 10 m/s2
Cercatu: Work done by the force of gravity (W)
Soluzione:
W = F d = w h = m g h
W = (1)(10)(2) = 20 Joule
W = work, F = force, d = distance, w = misfattu, 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
Cunnisciutu:
Massa (m) = 1 kg
Accelerazione di gravità (g) = 10 m/s2
Elongation (x) = 2 cm = 0.02 m
Pesu (w) = mg = (1 kg)(10 m/s2) = 10 kg m/s2 = 10 N
Cercatu: Spring constant and work done by spring force
Soluzione:
(a) Spring constant
Formula di Lege di Hooke :
F = k x.
k = F / x = w / x = m g / x
k = (1)(10) / 0.02 = 10 / 0.02
k = 500 N/m
(b) work done by spring force
W = – ½ k x2
W = – ½ (500)(0.02)2
W = – (250)(0.0004)
W = -0.1 Joule
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 forza di attritu Fk = 2 N. Determine the net work done on the box.
Cunnisciutu:
Forza (F) = 10 N
Force of kinetic friction (Fk) = 2 N
Displacement (d) = 2 m
Cercatu: Net work (Writi)
Soluzione:
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 :
Writi = W1 - W2
Writi = 20 - 4
Writi = 16 Joule
6. What is the work done by force F on the block.
Cunnisciutu:
Force (F) = 12 Newton
Displacement (d) = 4 meters
Vulutu: Work (W)
Soluzione:
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?
Cunnisciutu:
Force (F) = 200 Newton
Displacement (d) = 2 meters
Vulutu: Work (W)
Soluzione:
Travagliu:
W = F s
W = (200 Newton)(2 meters)
W = 400 N m
W = 400 Joule
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?
Cunnisciutu:
Displacement (d) = 10 meters – 0.5 meters = 9.5 meters
Force (F) = 50 Newton
Cercatu: Work (W)
Soluzione:
W = F s
W = (50 Newton)(9.5 meters)
W = 475 N m
W = 475 Joule
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.
Cunnisciutu:
Displacement (s) = 4 meters
Net force (F) = 50 Newton + 70 Newton = 120 Newton
Vulutu: Work (W)
Soluzione:
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?
Cunnisciutu:
Force (F) = 250 Newton
Displacement (s) = 1000 cm = 1000/100 meters = 10 meters
Cercatu: Work (W)
Soluzione:
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.
Cunnisciutu:
travagliu (W) = 375 Joule
Forza netta (ΣF) = 40 N + 10 N – 25 N = 25 Newton (diritta)
Cercatu: Sustantimentu (d)
Soluzione:
The equation of work :
W = F s
Object’s displacement :
d = W / F = 375 Joule / 25 Newton
d = 15 metris
12. The activities below which ùn fà travagliu 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
Soluzione:
The equation of work :
W = ΣF s
W= travagliu, F = forza, d = spustamentu
Based on the above formula, work done by force and there is a displacement.
A risposta curretta hè 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 muvimentu circulariu.
A. 0 Joule
B. 1400 Joule
C. 1540 Joule
D. 1760 Joule
Soluzione:
If the person pushes sedia à rotelle 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.
A risposta curretta hè 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
C. 1680 J
D. 2580 J
Cunnisciutu:
The force of push (F) = 350 Newton
Friction force (Fsoldi) = 70 Newton
Displacement of object (s) = 6 meters
Vulutu: Work (W)
Soluzione:
There are two forces that act on the object, the push force (F) and friction force (Fsoldi). 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 = – (Fsoldi)(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule
The net work :
W net = 2100 Joule – 420 Joule
W net = 1680 Joule
A risposta curretta hè 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 Joule
B. 3 Joule
C. 32 Joule
D. 192 Joule
Cunnisciutu:
Push force (F) = 14 Newton
Friction force (Fsoldi) = 10 Newton
Displacement of object (d) = 8 meters
Vulutu: Work (W)
Soluzione:
There are two forces that act on an object, push force (F) and friction force (Fsoldi).
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 = – (Fsoldi)(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule
The net work :
W net = 112 Joule – 80 Joule
W net = 32 Joule
A risposta curretta hè C.
16. Determine the net work based on figure below.
A. 360 Joule
B. 450 Joule
C. 600 Joule
D. 750 Joule
Soluzione:
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
A risposta curretta hè 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
Cunnisciutu:
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
Cercatu: The wood will enter entirely into the ground after…. hits.
Soluzione:
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
A risposta curretta hè D.
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