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Work done by a constant force is proportional to the force applied multiplied by the object's displacement. Forces acting across a distance are considered to have done work on an object. The change in kinetic energy that an item experiences is work done on it physically. Gaspard-Gustave Coriolis described it as "weight lifted through a height," based on the usage of early steam engines to move buckets of water out of flooded ore mines. The newton-meter or joule is the SI unit of work (J).
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Key Terms: Work Done, Force, Distance, Displacement, kinetic energy
Work Done by Constant Force: Definition
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Work done is defined as the product of the force and the distance travelled in the direction of the force when the force is constant.
Assume a body is held on a frictionless surface and is subjected to a force of constant magnitude of 10 N. The body will finish a distance of 5 metres due to the action of forces, and the work done will be stated as
W=Fs=10×5=50 Joule
Units
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Dimensional analysis is one method to verify if an expression is accurate. We've learned that work is defined as a change in an object's kinetic energy multiplied by the force times the distance. Both of these units should be the same. The units of kinetic energy — and other types of energy – are joules (J). Similarly, the force is measured in newtons (N) and the distance is measured in metres (m). The two assertions should be equal to one another if they are equivalent.
N⋅m=kgms2⋅m=kgm2s2=J
Positive, Negative and Zero work done
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- Positive Work: The work done is positive if a force displaces an item in its direction. The motion of a ball descending towards the floor, with the ball's displacement in the direction of gravity, is an example of this type of work.
- Negative Work: The work is said to be negative when the force and displacement are in opposing directions. For example, if a ball is thrown upwards, its displacement will be upwards, while the force due to gravity will be downwards.
- Zero Work: The work done by the force on the item is 0 if the force and displacement directions are perpendicular to each other.
When we push forcefully against a wall, for example, the force we are putting on the wall produces no work since the wall's displacement equals d = 0. Our muscles, on the other hand, use our own energy in this process, and as a result, we become tired.
Force at an Angle to Displacement
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We have assumed that every force exerted on an item is parallel to the direction of motion up until now. We've assumed that our motion is one-dimensional, acting just along the x and y axes. To better analyse and comprehend how nature operates in our three-dimensional world, we'll start by talking about how things work in two dimensions.
A force does not have to operate on an item parallel to the direction of motion, and it seldom does. We already deduced that W = F d, implying that the work done on an item is equal to the force acting on it multiplied by the displacement. This is not the end of the story, however. This formula includes an assumed cosine element, which we ignore for forces acting in the same direction as the motion. The cosine term equals one and does not affect the formula if the angle of the force along the direction of motion is zero, i.e. the force is parallel to the direction of motion.
As the angle of the force with regard to the direction of motion is increased, less and less work is done in the current direction, and more and more work is done in the perpendicular direction. This operation is continued until we are perpendicular to our original direction of motion, at which point the angle is 90 degrees and the cosine term equals zero, implying that no work is done in that direction. Instead, we're working in a different direction.
Work is the integral of the force and the dot product with respect to x, as we've seen. The dot product of force and a very tiny distance, on the other hand, is equal to the two terms times the cosine of the angle between them. F * dx = Fdcos(theta). Explicitly,
Angle: Both the force and the direction of motion are vectors, as you may recall. The cosine term is zero when the angle is 90 degrees. They add up to one when travelling in the same direction.
Examples
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Consider a coordinate system in which the abscissa is x and the ordinate is y. Consider a box that is being pushed in the x-direction. What happens in the three scenarios below?
- Is the box being moved in the same direction as the x-axis?
- Is the box being pushed at a 45-degree angle to the x-axis?
- Is the box being pushed at a 60-degree angle to the x-axis?
- Is the box being pushed at a 90-degree angle to the x-axis?
- We know that in the first situation, all of the force is acting on the box in the x-direction, hence work will only be done in that direction. Furthermore, the box is not moving in the vertical direction — it remains unchanged in the Y-direction. The angle is equal to zero since the force is operating parallel to the direction of motion, and our total effort is simply the force times the displacement in the x-direction.
- The box is pushed at a 45-degree angle to the x-direction in the second situation, and consequently at a 45-degree angle to the y-direction as well. The cosine of 45 degrees equals 1/2, or about 0.71 when calculated. This suggests that 71 percent of the force is involved in the effort in the x-direction. The remaining 29% is operating in the y-direction.
- We know that the force is operating at a 60-degree angle to the x-direction in the third situation, and hence at a 30-degree angle to the y-direction. The cosine of 60 degrees is equal to 1/2 when assessed. This indicates that the force is operating in both the x and y directions! In terms of the work done, it is linear.
- The box is moved at an angle perpendicular to the x-axis in the last scenario. To put it another way, we're pushing the box in a Y-direction! As a result, the box's location will remain static and will not move along the x-axis. The amount of work done in the direction of the x-axis will be 0.
Things to Remember
- Work is a unit of measurement for the amount of energy consumed in moving an item; the most common formula is force times displacement. If the thing does not move, no work is done.
- Work is a change in kinetic energy in a system.
- Displacement is not the same as distance. The total displacement of a box pushed 3 metres ahead and then 4 metres to the left is 5 metres, not 7 metres.
- The force times the displacement times the cosine of the angle equals the work done on an item in a particular direction of motion.
- There is no work done if the force is perpendicular to the direction of motion.
- We ignore the cosine element when evaluating force parallel to the direction of motion since it equals 1, hence it does not change the expression.
Sample Questions
Ques 1. Explain Positive, Negative and Zero Work. Give one example of each. (4 Marks)
Ans: POSITIVE WORK: When force and displacement are in the same direction, the work done on an item is said to be positive work.
Example: Force and displacement act in the same direction when an item travels on a horizontal surface. As a result, the work completed is favourable.
NEGATIVE WORK: When force and displacement are in opposing directions, the work done is considered to be negative work.
Example: When an item is tossed upwards, gravity operates downwards and displacement acts above.
ZERO WORK: When force and displacement are perpendicular to one other or when either force or displacement is zero, the work done is said to be zero.
When we walk with an item in our hands, the force operates in a downward direction, while displacement acts in a forward direction.
Ques 2. Convert 1 joule to ergs. (5 Marks)
Ans: Joule: SI system, erg: CGS system
Work = Force x Distance = Mass x Acceleration x Length = Mass x Length/(Time)^2 ×Length
1 Joule of Energy = 1 Newton x 1 metre
1 erg = 1 dyne x 1 cm
1 joule= 1 newton x 1 metre
1 joule = 105 dynes x 102 cm
1 joule =107 erg
Dimension Formula for Joule is [M¹L²T?²]
Let N1(joule) = N2(erg)
N1[M?¹L?²T??²] = N2[M?¹L?²T??²]
N2 = N1[M?¹L?²T??²] /[M?¹L?²T??²]
N1 x [M1/M2]¹ x [L1/L2]² x [T1/T2]?²
1 x [1kg/gm]¹ x [1m/1cm]² x [1s/1s]?²
1 x (1000g/gm) x (100cm/1cm)² x (1s/1s)?²
⇒ 10?
N1(joule) = N2(erg)
∴ 1 Joule = 10?erg
Ques 3. A force of 7 N acts on an object. The displacement is 8 m, in the direction of the force. Consider forces acting on the object through displacement. What is the work done in this case? (2 Marks)
Ans: Given, force=7N; displacement=8m
Workdone=Force×Displacement
=7×8=56J
Ques 4. Convert 1 kWh into joules. (2 Marks)
Ans: 1kWh=1kW×1h
=1000W×3600s
=3600000J
=3.6×10^6 J
Ques 5. A block weighing 20 N is lifted 6 m vertically upward. Calculate the potential energy stored in it. (3 Marks)
Ans: Weight of the block = W = 20 N
Height = h = 6 m
Potential energy = P. E. =?
P.E. = mgh
We know that w = mg
P.E. = (mg) x h
Thus, P.E. = (2 x 10) x 6
P.E. = 120 J
Ques 6. A car weighing 12 k N has a speed of 20 ms –1. Find its kinetic energy. (240 kJ) (3 Marks)
Ans: Weight of the car = w = 12kN = 12 x 1000 N = 12000 N
Speed of the car = v = 20 ms^– 1
Kinetic energy = K.E. =?
K.E. =1/2 m v^2
W = mg or m =1200 kg
Thus K.E. = 0.5 x 1200 x (20)^2
= 600 X 400 = 240000 J
K.E. = 240 kJ
Ques 7. A 50 kg man moved 25 steps up in 20 seconds. Find his power, if each step is 16 cm high. (3 Marks)
Ans: Mass = m = 50 kg
Total height = h = 25 x 16 = 400 cm = 400/100 m = 4 m
Time = t = 20 s
Power = P = ?
Power = work/time = w/t = mgh/t
P = 50*10*4/20
P = 100 W
Ques 8. 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? (2 Marks)
Ans: Force (F) = 10 N
The horizontal force (Fx) = F cos30 = (10)(0.5√3) = 5√3 N
Displacement (d) = 1 meter
W = Fx d = (5√3)(1) = 5√3 Joule
Ques 9. 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. (2 Marks)
Ans:
Known :Force (F) = 20 N
Displacement (s) = 2 m
Angle (θ) = 0
Work (W)= F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule
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