UNISA
Module Code: PHY1505
Module Title: Elementary Mechanics
Assessment Format: Take-Home Assignment
Total Marks: 80
Duration: 03 -14 August 2026
First Examiner: Dr SJ Mofokeng
Second Examiner: Dr M Ramantswana
,QUESTION 1
Given
• Mass:
𝑚 = 3.37 × 105 g = 337 kg
• Distance lowered:
𝑑 = 600 cm = 6.0 m
• Tension 1:
𝑇1 = 1830 N
• Tension 2:
𝑇2 = 1295 N
• Weight:
𝐹𝐺 = 2500 N
The object moves vertically downward by 6 m.
The work done by a force is
𝑊 = 𝐹𝑑 cos 𝜃
where
• 𝐹= force
• 𝑑= displacement
• 𝜃= angle between the force and displacement.
1.1 Work done by 𝑻𝟏
The rope is 60° above the horizontal.
Therefore the angle between the tension and the downward displacement is
𝜃 = 150∘
Substitute into the work formula:
𝑊 = 𝑇1 𝑑 cos 150∘
= ሺ1830ሻሺ6ሻ cos 150∘
, Since
cos 150∘ = −0.866
𝑊 = ሺ1830ሻሺ6ሻሺ−0.866ሻ
𝑊 = −9508.7 J
Answer
𝑊𝑇1 = −9.51 × 103 J
1.2 Work done by 𝑻𝟐
The rope is 45° above the horizontal.
Angle between the force and downward motion:
𝜃 = 135∘
Using
𝑊 = 𝐹𝑑 cos 𝜃
= ሺ1295ሻሺ6ሻ cos 135∘
Since
cos 135∘ = −0.707
= ሺ1295ሻሺ6ሻሺ−0.707ሻ
= −5493 J
Answer
𝑊𝑇2 = −5.49 × 103 J
1.3 Work done by Gravity
Gravity acts in the same direction as the displacement.
Therefore
𝜃 = 0∘
𝑊 = 𝐹𝑑 cos 0∘
= ሺ2500ሻሺ6ሻሺ1ሻ
= 15000 J
Answer 𝑊𝐺 = 1.50 × 104 J