Calculating Work on a Box at an Angle

What happens when the applied force and the motion are not parallel? By using the formula in this lesson, we are actually calculating the component of the applied force in the direction of motion. Note that the component of the force perpendicular to the ...

What happens when the applied force and the motion are not parallel? By using the formula \(W=F\Delta x\cos\theta\) , we are actually calculating the component of the applied force in the direction of motion. Note that the component of the force perpendicular to the direction of motion does no work.

Example: Calculating Work Done on a Box Pulled at an Angle

Question

Calculate the work done on a box, if it is pulled \(\text{5}\) \(\text{m}\) along the ground by applying a force of \(F=\text{20}\text{ N}\) at an angle of \(\text{60}\)\(\text{°}\) to the horizontal.

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Step 1: Analyse the question to determine what information is provided

  • The force applied is \(F\)=\(\text{20}\) \(\text{N}\)

  • The distance moved is \(\Delta x\) = \(\text{5}\) \(\text{m}\) along the ground

  • The angle between the applied force and the motion is \(\theta\)=\(\text{60}\)\(\text{°}\)

These quantities are in the correct units so we do not need to performany unit conversions.

Step 2: Analyse the question to determine what is being asked

  • We are asked to find the work done on the box.

Step 3: Substitute and calculate the work done

Now we can calculate the work done on the box:

\begin{align*} W& = {F}\Delta x\cos\theta\\ & = \left(20 \right)\left(5 \right) \left(\cos 60\right)\\ & = \text{50}\text{ J} \end{align*}

Note that the answer is positive as the component of the force parallel to the direction of motion is in the same direction as the motion.

The work done on the box is \(\text{50}\) \(\text{J}\).

This lesson is part of:

Work, Energy and Power

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