Work

- Work done by a constant force is given by the product of the force and the distance moved in the direction of the force.
- The unit of Nm(Newton metre) or J(Joule).
- Work is a scalar quantity.

When the direction of force and motion are same, θ = 0^{o}, therefore cosθ = 1

Work done,

W = F × s

Example 1

A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force.

**Answer:**

Work done,

W = F × s × cos θ

W = 50 × 3.0 × cos30^{o} = 129.9J

Example 2

Diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find

- the work done by the pulling force
- the work done by the frictional force

**Asnwer:**

(a) The force is in the same direction of the motion. Work done by the pulling force,

W = F × s = (10)(2) = 20J

(b) The force is not in the same direction of motion, work done by the frictional force

W = F × s × cos180o= (5)(2)(-1) = -10J

Work Done Against the Force of Gravity

Example 3

Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's mass is 45kg, find the work done by Ranjit to reach the top of the staircase.

**Answer**:

In this case, Ranjit does work to overcome the gravity.

Ranjit's mass = 45kg

Vertical height of the motion, h = 35 × 0.15

Gravitational field strength, g = 10 ms-2

Work done, W = ?

W = mgh = (45)(10)(35 × 0.15) = 2362.5J

Finding Work from Force-Displacement Graph

In a Force-Displacement graph, work done is equal to the

**area**in between the graph and the horizontal axis.

Example 4

The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley.

**Answer**:

In a Force-Displacement graph, work done is equal to the area below the graph. Therefore, work done

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