Moderate -0.3 This is a straightforward work-energy problem requiring standard mechanics techniques: resolving forces on an inclined plane, calculating friction, applying constant speed condition (F=0), then computing work done. The multi-step nature and need to handle trigonometry with tan θ = 5/12 makes it slightly below average difficulty, but it's still a routine textbook-style question with no novel insight required.
A particle of mass 13 kg is on a rough plane inclined at an angle of \(\theta\) to the horizontal, where \(\tan \theta = \frac{5}{12}\). The coefficient of friction between the particle and the plane is 0.3. A force of magnitude \(T\) N, acting parallel to a line of greatest slope, moves the particle a distance of 2.5 m up the plane at a constant speed. Find the work done by this force. [5]
Apply Newton’s second law parallel to the plane with a = 0
WD = T × 2.5 [= 86 × 2.5]
M1
WD = T × d
WD = 215 J
A1
Alternative method for question 3
Answer
Marks
Guidance
R = 13g cos 22.6 = 13g × (12/13), [R = 120]
B1
Resolve perpendicular to the plane
F = 0.3 × 13g cos 22.6 [F = 36]
M1
Using F = µR
PE gain = 13 × g × 2.5 × (5/13) [= 125]
M1
Attempt PE gain. Allow sin 22.6 for 5/13
[WD by T = 13 × g × 2.5 × (5/13) + F × 2.5]
M1
Using WD by T = PE gain + WD against F
WD by T = 215 J
A1
5
Answer
Marks
Guidance
Question
Answer
Marks
Question 3:
3 | R = 13g cos 22.6 = 13g × (12/13), [R = 120] | B1 | Resolve perpendicular to the plane
F = 0.3 × 13g cos 22.6 [F = 36] | M1 | Using F = µR
T = F + 13g sin 22.6 = F + 13g × (5/13), [T = 86] | M1 | Apply Newton’s second law parallel to the plane with a = 0
WD = T × 2.5 [= 86 × 2.5] | M1 | WD = T × d
WD = 215 J | A1
Alternative method for question 3
R = 13g cos 22.6 = 13g × (12/13), [R = 120] | B1 | Resolve perpendicular to the plane
F = 0.3 × 13g cos 22.6 [F = 36] | M1 | Using F = µR
PE gain = 13 × g × 2.5 × (5/13) [= 125] | M1 | Attempt PE gain. Allow sin 22.6 for 5/13
[WD by T = 13 × g × 2.5 × (5/13) + F × 2.5] | M1 | Using WD by T = PE gain + WD against F
WD by T = 215 J | A1
5
Question | Answer | Marks | Guidance
A particle of mass 13 kg is on a rough plane inclined at an angle of $\theta$ to the horizontal, where $\tan \theta = \frac{5}{12}$. The coefficient of friction between the particle and the plane is 0.3. A force of magnitude $T$ N, acting parallel to a line of greatest slope, moves the particle a distance of 2.5 m up the plane at a constant speed. Find the work done by this force. [5]
\hfill \mbox{\textit{CAIE M1 2019 Q3 [5]}}