(a) What is work? (b) How do we determine the amount of work done, given the force associated with the work?

Short Answer

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(a) Work is a scalar quantity representing the energy transferred to an object when it is moved in the direction of an applied force. It is calculated as the product of the force and the distance the object moves, accounting only for the component of the force acting along the direction of displacement. (b) To determine the amount of work done (W), given the force (F) and distance (d), we use the formula: \(W = Fd\cos(\theta)\), where \(\theta\) is the angle between the direction of the force and the direction of displacement. If the force is applied in the same direction as the displacement (\(\theta = 0\)), the formula becomes: \(W = Fd\).

Step by step solution

01

(a) Definition of Work

Work is a measure of the energy transferred to an object when it is moved in the direction of the applied force. In physics, work is a scalar quantity that is calculated as the product of the force applied to an object and the distance it moves as a result of that force, only accounting for the component of the force that is acting along the direction of displacement.
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(b) Determining the Amount of Work Done

To determine the amount of work done on an object given the force F associated with the work, we need two key variables: \(F\) (the force applied in Newtons, often represented as N) and \(d\) (the distance the object moves in meters, represented as m). The formula used to calculate work (W) is: \[W = Fd\cos(\theta)\] Where \(W\) represents the work done in Joules (J), \(\theta\) represents the angle between the direction of the force and the direction of displacement. If the force is applied in the same direction as the displacement, \(\theta = 0\), then the formula becomes: \[W = Fd\] For example, if a force of 10 N is applied to an object and the object moves 5 meters in the direction of the force (so \(\theta = 0\)), The work done would be: \[W = (10 \, N)(5 \, m) = 50 \, J\] Thus, the amount of work done is 50 Joules.

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