CAIE M1 2005 June — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2005
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
DifficultyModerate -0.8 This is a straightforward work-energy question requiring only the formula W = Fs cos θ and basic algebraic manipulation. Students calculate distance (30m) from speed and time, then solve 720 = 30 × 30 × cos θ to find θ = 53.1°. No problem-solving insight needed, just direct application of a standard formula.
Spec6.02a Work done: concept and definition

A small block is pulled along a rough horizontal floor at a constant speed of \(1.5 \text{ m s}^{-1}\) by a constant force of magnitude \(30 \text{ N}\) acting at an angle of \(\theta°\) upwards from the horizontal. Given that the work done by the force in \(20 \text{ s}\) is \(720 \text{ J}\), calculate the value of \(\theta\). [3]

Question 1:
AnswerMarks
1× θ
720 = 30(1.5 20)cos
θ
AnswerMarks
= 36.9M1
A1
AnswerMarks
A1 3α
For using WD = Fdcos
or P = WD/T and P = (Fcos α )v
Question 1:
1 | × θ
720 = 30(1.5 20)cos
θ
= 36.9 | M1
A1
A1 3 | α
For using WD = Fdcos
or P = WD/T and P = (Fcos α )v
A small block is pulled along a rough horizontal floor at a constant speed of $1.5 \text{ m s}^{-1}$ by a constant force of magnitude $30 \text{ N}$ acting at an angle of $\theta°$ upwards from the horizontal. Given that the work done by the force in $20 \text{ s}$ is $720 \text{ J}$, calculate the value of $\theta$.
[3]

\hfill \mbox{\textit{CAIE M1 2005 Q1 [3]}}