IB Physics SL A.3.4 Work by Constant Force

Practise calculating work from the force component along displacement, recognising zero work for perpendicular forces and linking positive or negative work to energy change.

Exam points

calculate W = Fs cos θ with θ between the force and displacement, not another diagram angle
state that a centripetal or other perpendicular force does zero work because cos 90° = 0
use signed work by friction, weight or an applied force to determine a distance or energy change

[Maximum number: 2]

This question is in two parts. Part 1 is about two children on a merry-go-round. Part 2 is about electric circuits.

Part 1 Two children on a merry-go-round
Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Figure for Question A.3.4—Work by constant force question 1 — IB Physics SL

Aibhe is moving at speed $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

Euan drags his foot along the ground to act as a brake. The merry-go-round comes to a stop after 4.0 rotations. The radius of the merry-go-round is 1.5 m . The average frictional force between his foot and the ground is 45 N . Calculate the work done.

distance travelled by Euan $=4.0 \times 2 \pi \times 1.5(=37.70 \mathrm{~m})$ ;

W\left(=F_{a v} d=45 \times 37.70\right)=1700(\mathrm{~J}) ;

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A.3.4—Work by constant force question 1