Lesson 7, Topic 1
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# The Concept of Work

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When a man moves from one point to another carrying a load, the man has performed work, but when a boy carries a load for 2 hours without moving from one point to another, no work has been done. Therefore, for work to be done, force must be applied to a body through a particular distance.

Therefore, work is said to be done whenever a force moves a body through a distance in the direction of the force. Or Work is defined as the product of the force and its perpendicular distance moved in the direction of the force.

Work done = force x distance in the direction of the force

= force x displacement

= Newton x metre

= Nm

Also, Remember that Force = mass x acceleration

Therefore, Work done can also be written as mass x acceleration x displacement

The S.I unit of work is joule (J) or Newton metre (Nm)

Example

i. A man applied a force of 20N to a cart which moved through a distance of 15m. Calculate the work done by the man.

Solution:

F = 20N, distance = 15m

Work done W = Force F x dsitance d

= 20N x 15M

= 300Joules.

ii.  If the work done by a force on an object is 30J and the distance moved by the object is 1.45m, calculate the amount of force that was applied on the object.

Solution:

Work done = 30J, distance = 1.45m

Work done W = Force F x dsitance d

Force = $$\frac {Work-done}{distance}$$

Force = $$\frac {30J}{1.45m}$$

Force = 20.67N

iii.  Calculate the amount of work that is done when a force of 35N moves a 12kg mass through a distance of 1.25 m.

Solution:

mass = 12kg, force = 35N

distance = 1.125

Formula for work done = Force x distance

= 35 x 1.25

= 43.75 J

When a force F is applied at an inclined angle θ to a body at an angle θ

Work done = F x dθ

= Fdcosθ

When a body is lifted to a height, h, work is done against gravity, hence the work done,

w = weight of the body x height

w = mg x h

w = mgh

Where,

m = mass of the body

g = acceleration due to gravity

h = height to which the body is raised

Example:

1. A boy of mass 80kg runs up a staircase of 30 steps each of 15cm.

Calculate the work done by the boy. (g = 10ms-2).

Solution

Work done = Force x distance

F = mg = 80 x 10 = 800N

m = 15cm, convert to metre = $$\frac{15}{100} \scriptsize m$$

Distance = $$\left(\frac{15}{100} \right) \scriptsize m \: \times \: 30$$

Distance = 0.15 x 30

= 4.5m

Work done = mg x h

= 800 x 4.5

= 3600 Joules.

2. A force of 50N is inclined at an angle of 30° makes a wheelbarrow to move through a distance of 40m. Calculate the work done.

Solution

Work done = Force x distance

distance = dcosθ  = 40cos30°

= 40 x 0.8660

Work done = Force x distance

= 50 x 40 x 0.8660

= 1732 J

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