JSS2 Basic Science Lesson Note on Calculation involving Workdone

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TOPIC: WORK, ENERGY AND POWER 2

Content: Calculation involving work done.

1. What work is done when a mass of 5.00kg is raised through a vertical height of 2.5m (acceleration due to gravity is 10m/sec)?

Solution

Work-done =mass acceleration  distance

Since force= mass  acceleration

W (d) =mass accel. distance

=5.00kg  10m/s²  2.5m

=125joules

2. Calculate the work-done by a body of weight 200N which, moves through a distance of 12N

Solution

Force=200N

Distance moved=12m

Work-done=force distance moved

=(200 12)joules

=2400joules.

3. Find the work done by a man who lifts a weight of 20N from the ground to a vertical height of 4m by means of a rope.

Solution:

W =? F = 20N, S = 4m

W = FXS

W =20x 4= 80Joules (80J).

4. A boy pushes a box along a horizontal surface with a force of 15N. Find the work done in moving the box through a distance of 5m.

Solution:

W =? F = 15N, S =5m

W = F X S

W = 15 x 5 = 75J.

5. A pumping machine raise 75kg of water through a height of 5m, what is the work done by the machine? (g = 10m/s²).

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