In this question, use g=9.8 m s−2g = 9.8 \text{ m s}^{-2}g=9.8 m s−2.
A precision sensor module is dragged across a rough horizontal laboratory floor. A horizontal tension of magnitude FFF newtons is applied to the module by a robotic arm.
The module moves with a constant acceleration of 0.75 m s−20.75 \text{ m s}^{-2}0.75 m s−2.
The sensor module has a weight of 4.41 N4.41 \text{ N}4.41 N.
The only resistance to motion is the friction between the module and the floor.
The coefficient of friction between the module and the floor is 0.220.220.22.
Find FFF.
87 exam-style questions on OCR (MEI) A Level Maths 3.4 Forces, covering 3.4.1 Language relating to forces, 3.4.2 Acceleration due to gravity, 3.4.3 Identify forces and draw force diagrams, 3.4.4 Resultant of concurrent forces, 3.4.5 Equilibrium of a particle, 3.4.6 Resolve a force into components (A-level only), 3.4.7 Equilibrium condition for resultant zero (A-level only), 3.4.8 Forces in equilibrium sum to zero (A-level only), 3.4.9 Equations for a particle in equilibrium (A-level only), 3.4.10 Frictional force and normal contact force (A-level only), 3.4.11 Model friction with F ≤ μR (A-level only), and 3.4.12 Apply Newton's Laws to friction problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.