AQA M2 2013 January — Question 3 5 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
TopicPower and driving force
TypeFind power at constant speed
DifficultyModerate -0.3 This is a straightforward application of the work-energy principle at constant speed. Students need to recognize that at constant speed, driving force equals resistance plus component of weight down the slope, then apply P = Fv. The calculation is direct with no conceptual subtlety, making it slightly easier than average but still requiring proper setup of forces on an incline.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

3 A van, of mass 1500 kg , travels at a constant speed of \(22 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) up a slope inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 25 }\). The van experiences a resistance force of 8000 N .
Find the power output of the van's engine, giving your answer in kilowatts.

3 A van, of mass 1500 kg , travels at a constant speed of $22 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a slope inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 25 }$.

The van experiences a resistance force of 8000 N .\\
Find the power output of the van's engine, giving your answer in kilowatts.

\hfill \mbox{\textit{AQA M2 2013 Q3 [5]}}