Work done by force – problems and solutions

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.

Work done by a force – problems and solutions 1

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

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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!

Work done by a force – problems and solutions 2

Known :

Force (F) = 10 N

The horizontal force (Fx) = F cos 30o = (10)(0.5√3) = 5√3 N

Displacement (d) = 1 meter

Wanted : Work (W) ?

Solution :

W = Fx 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/s2, 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/s2

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.

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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

Known :

Mass (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s2

Elongation (x) = 2 cm = 0.02 m

Weight (w) = m g = (1 kg)(10 m/s2) = 10 kg m/s2 = 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 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.

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

Work done by a force – problems and solutions 3

Known :

Force (F) = 10 N

Force of kinetic friction (Fk) = 2 N

Displacement (d) = 2 m

Wanted : Net work (Wnet)

Solution :

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 :

Wnet = W1 – W2

Wnet = 20 – 4

Wnet = 16 Joule

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

Known :Work done by force – problems and solutions 1

Force (F) = 12 Newton

Displacement (d) = 4 meters

Wanted: Work (W)

Solution :

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

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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 :Work done by force – problems and solutions 2

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

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9.

Work done by force – problems and solutions 3

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 :Work done by force – problems and solutions 4

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.

Work done by force – problems and solutions 11

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 (Ffric) = 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 (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

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 (Ffric) = 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 (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

The correct answer is C.

16. Determine the net work based on figure below.

A. 360 JouleWork

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 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

Known :

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

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.

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