Moderate -0.5 This is a straightforward work-energy application requiring only the work-energy principle (work done by pulling force = work against resistance + gain in PE) and the formula work = force × distance × cos(angle). No problem-solving insight needed, just direct substitution into standard formulas.
\includegraphics{figure_2}
A crate \(C\) is pulled at constant speed up a straight inclined path by a constant force of magnitude \(F\) N, acting upwards at an angle of 15° to the path. \(C\) passes through points \(P\) and \(Q\) which are 100 m apart (see diagram). As \(C\) travels from \(P\) to \(Q\) the work done against the resistance to \(C\)'s motion is 900 J, and the gain in \(C\)'s potential energy is 2100 J. Write down the work done by the pulling force as \(C\) travels from \(P\) to \(Q\), and hence find the value of \(F\). [3]
\includegraphics{figure_2}
A crate $C$ is pulled at constant speed up a straight inclined path by a constant force of magnitude $F$ N, acting upwards at an angle of 15° to the path. $C$ passes through points $P$ and $Q$ which are 100 m apart (see diagram). As $C$ travels from $P$ to $Q$ the work done against the resistance to $C$'s motion is 900 J, and the gain in $C$'s potential energy is 2100 J. Write down the work done by the pulling force as $C$ travels from $P$ to $Q$, and hence find the value of $F$. [3]
\hfill \mbox{\textit{CAIE M1 2009 Q2 [3]}}