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.

__Known :__

Force (F) = 20 N

Displacement (s) = 2 m

Angle (θ) = 0

__Wanted__ : Work (W)

__Solution :__

W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule

[irp]

2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30^{o} angle as shown in figure below. Determine the work done by force F!

Known :

Force (F) = 10 N

The horizontal force (F_{x}) = F cos 30^{o} = (10)(0.5√3) = 5√3 N

Displacement (d) = 1 meter

__Wanted__ : Work (W) ?

__Solution__ :

W = F_{x} d = (5√3)(1) = 5√3 Joule

3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s^{2}, determine the work done by the force of gravity!

__Known :__

Object’s mass (m) = 1 kg

Height (h) = 2 m

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Work done by the force of gravity (W)

__Solution :__

W = F d = w h = m g h

W = (1)(10)(2) = 20 Joule

*W = work, F = force, d = distance, w = weight, h = height, m = mass, g = acceleration due to gravity.*

[irp]

4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m/s^{2}, determine (a) the spring constant (b) work done by spring force on object

__Known :__

Mass (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Elongation (x) = 2 cm = 0.02 m

Weight (w) = m g = (1 kg)(10 m/s^{2}) = 10 kg m/s^{2 }= 10 N

__Wanted :__ Spring constant and work done by spring force

__Solution :__

__(a) Spring constant__

**Formula of Hooke’s law **:

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 x^{2}

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.

[irp]

5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force F_{k }= 2 N. Determine the net work done on the box.

__Known :__

Force (F) = 10 N

Force of kinetic friction (F_{k}) = 2 N

Displacement (d) = 2 m

__Wanted :__ Net work (W_{net})

__Solution :__

Work done by force F :

W_{1} = F d cos 0 = (10)(2)(1) = 20 Joule

Work done by force of kinetic friction (F_{k}) :

W_{2} = F_{k} d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule

Net work :

W_{net} = W_{1} – W_{2}

W_{net} = 20 – 4

W_{net} = 16 Joule

6. What is the work done by force F on the block.

__Known :__

Force (F) = 12 Newton

Displacement (d) = 4 meters

__Wanted:__ Work (W)

__Solution :__

W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule

[irp]

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?

__Known :__

Force (F) = 200 Newton

Displacement (d) = 2 meters

__Wanted:__ Work (W)

__Solution :__

Work :

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?

__Known :__

Displacement (d) = 10 meters – 0.5 meters = 9.5 meters

Force (F) = 50 Newton

__Wanted :__ Work (W)

__Solution :__

W = F s

W = (50 Newton)(9.5 meters)

W = 475 N m

W = 475 Joule

[irp]

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.

__Known :__

Displacement (s) = 4 meters

Net force (F) = 50 Newton + 70 Newton = 120 Newton

__Wanted:__ Work (W)

__Solution :__

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?

__Known :__

Force (F) = 250 Newton

Displacement (s) = 1000 cm = 1000/100 meters = 10 meters

__Wanted :__ Work (W)

__Solution :__

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.

__Known :__

Work (W) = 375 Joule

Net force (ΣF) = 40 N + 10 N – 25 N = 25 Newton (rightward)

__Wanted :__ Displacement (d)

__Solution :__

The equation of work :

W = F s

Object’s displacement :

d = W / F = 375 Joule / 25 Newton

d = 15 meters

12. The activities below which do not do work 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

Solution :

The equation of work :

W = ΣF s

*W = **work**, F = **force**, **d** = **displacement*

Based on the above formula, work done by force and there is a displacement.

The correct answer is 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 circular motion.

A. 0 Joule

B. 1400 Joule

C. 1540 Joule

D. 1760 Joule

Solution :

If the person pushes *wheelchair** 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.

The correct answer is 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

__Known :__

The force of push (F) = 350 Newton

Friction force (F_{fric}) = 70 Newton

Displacement of object (s) = 6 meters

__Wanted:__ Work (W)

__Solution :__

There are two forces that act on the object, the push force (F) and friction force (F_{fric}). 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 = – (F_{fric})(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule

__The net work :__

W net = 2100 Joule – 420 Joule

W net = 1680 Joule

The correct answer is 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

__Known :__

Push force (F) = 14 Newton

Friction force (F_{fric}) = 10 Newton

Displacement of object (d) = 8 meters

__Wanted:__ Work (W)

__Solution :__

There are two forces that act on an object, push force (F) and friction force (F_{fric}).

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 = – (F_{fric})(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule

__The net work :__

W net = 112 Joule – 80 Joule

W net = 32 Joule

The correct answer is C.

16. Determine the net work based on figure below.

A. 360 Joule

B. 450 Joule

C. 600 Joule

D. 750 Joule

Solution :

__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

The correct answer is 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 10^{3} N and the acceleration due to gravity is 10 m/s^{2}, then the wood will enter entirely into the ground after…. hits.

A. 4

B. 16

C. 28

D. 30

__Known :__

Mass of hammer (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s^{2}

*Weight of hammer (w) = m g = (10)(10) = 100 kg m/s*^{2}

Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters

The resistance of wood (F) = 2 x 10^{3} N = 2000 N

Length of wood (s) = 60 cm = 0.6 meters

__Wanted :__ The wood will enter entirely into the ground after…. hits.

__Solution :__

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

The correct answer is D.

[wpdm_package id=’1192′]

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