Question 1

Multiple choiceMedium1 mark

A constant force of magnitude 10.0 N10.0\ \text{N} acts on a box, directed 30.0∘30.0^\circ north of east. The box moves 6.0 m6.0\ \text{m} east and then 8.0 m8.0\ \text{m} north. What is the total work done by the force on the box?

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Answer: D. Work on each segment is the force component along that segment multiplied by its length. The east and north components are 10.0cos⁡30.0∘=8.66 N10.0\cos30.0^\circ=8.66\ \text{N} and 10.0sin⁡30.0∘=5.00 N10.0\sin30.0^\circ=5.00\ \text{N} Thus W=(8.66)(6.0)+(5.00)(8.0)=91.96 JW=(8.66)(6.0)+(5.00)(8.0)=91.96\ \text{J} which is 92 J92\ \text{J} to two significant figures. A: subtracts the northward contribution instead of adding it, giving 52−40=12 J52-40=12\ \text{J} B: includes only the work during the northward displacement, 40 J40\ \text{J} C: includes only the work during the eastward displacement, 52 J52\ \text{J} Remember: add the work contributions from successive displacements, including their signs.

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