Edexcel M1 — Question 3 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeSingle angled force - find limiting friction or coefficient
DifficultyStandard +0.3 This is a standard M1 friction problem requiring resolution of forces and application of F=μR, followed by using F=ma with constant acceleration. The steps are routine: resolve vertically to find R, use limiting friction for part (a), find acceleration from kinematics, then resolve horizontally with F=ma for part (b). While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average.
Spec3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

A string is attached to a packing case of mass 12 kg, which is at rest on a rough horizontal plane. When a force of magnitude 50 N is applied at the other end of the string, and the string makes an angle of 35° with the vertical as shown, the case is on the point of moving. \includegraphics{figure_3}
  1. Find the coefficient of friction between the case and the plane. [5 marks]
The force is now increased, with the string at the same angle, and the case starts to move along the plane with constant acceleration, reaching a speed of 2 ms\(^{-1}\) after 4 seconds.
  1. Find the magnitude of the new force. [5 marks]
  2. State any modelling assumptions you have made about the case and the string. [2 marks]

AnswerMarks
(a) Resolve: \(R + 50 \sin 35° = 12g\), \(50 \cos 35° = \mu R\)M1 A1 A1
\(\mu(12g - 50 \sin 35°) = 50 \cos 35°\)M1 A1
\(\mu = 0.41\)
(b) Resolve: \(R + F \sin 35° = 12g\), \(F \cos 35° - \mu R = 12a\)M1 A1
\(a = 0.5\): \(F(\cos 35° + 0.41 \sin 35°) = 6 + 0.461(12g)\)M1 A1
\(F = 55.5 \text{ N}\)B1
(c) Case \(=\) particle (does not topple); string light and inextensibleB1 B1
Total: 12 marks
(a) Resolve: $R + 50 \sin 35° = 12g$, $50 \cos 35° = \mu R$ | M1 A1 A1 |
| $\mu(12g - 50 \sin 35°) = 50 \cos 35°$ | M1 A1 |
| $\mu = 0.41$ | |

(b) Resolve: $R + F \sin 35° = 12g$, $F \cos 35° - \mu R = 12a$ | M1 A1 |
| $a = 0.5$: $F(\cos 35° + 0.41 \sin 35°) = 6 + 0.461(12g)$ | M1 A1 |
| $F = 55.5 \text{ N}$ | B1 |

(c) Case $=$ particle (does not topple); string light and inextensible | B1 B1 |
| | **Total: 12 marks**
A string is attached to a packing case of mass 12 kg, which is at rest on a rough horizontal plane. When a force of magnitude 50 N is applied at the other end of the string, and the string makes an angle of 35° with the vertical as shown, the case is on the point of moving.

\includegraphics{figure_3}

\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of friction between the case and the plane. [5 marks]
\end{enumerate}

The force is now increased, with the string at the same angle, and the case starts to move along the plane with constant acceleration, reaching a speed of 2 ms$^{-1}$ after 4 seconds.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the magnitude of the new force. [5 marks]
\item State any modelling assumptions you have made about the case and the string. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q3 [12]}}