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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4Topics
    |
    1 Quiz
  2. Measurement | Week 2
    3Topics
  3. Measurement of Mass | Week 3
    6Topics
    |
    1 Quiz
  4. Motion | Week 4
    5Topics
    |
    1 Quiz
  5. Velocity-Time Graph | Week 5
    4Topics
    |
    1 Quiz
  6. Causes of Motion | Week 6
    5Topics
    |
    1 Quiz
  7. Work, Energy & Power | Week 7
    3Topics
  8. Energy Transformation / Power | Week 8
    3Topics
    |
    1 Quiz
  9. Heat Energy | Week 9
    5Topics
    |
    1 Quiz
  10. Linear Expansion | Week 10
    6Topics
    |
    1 Quiz
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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  

PHY13 e1600254766556

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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