A particle of mass 1.2kg is placed on a rough plane which is inclined at an angle θ to the horizontal, where sin θ = 3/25. The particle is kept in equilibrium by a horizontal force of magnitude P N acting in a vertical plane containing a line of greatest slope (see [Figure 6.1]). The coefficient of friction between the particle and the plane is 0.15. Find the least possible value of P.
- Find the least possible value of P.[6]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- *M1 Attempt at resolving perpendicular to the plane to get an equation
- A1 R=1.2g×(24/25)+P×(7/25)
- *M1 Attempt at resolving parallel to the plane to get an equation
- A1 F+P×(24/25) = 1.2g×(7/25)
- DM1 Use of F=0.15R to get an equation in P only
- A1 P=1.63
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →