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PHY1505 ASSIGNMENT 3 Solutions 2026

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PHY1505 ASSIGNMENT 3 Solutions 2026 UNISA 0-7-9-3-2-2-6-4-2-7 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

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PHY1505 ASSIGNMENT 3 Solutions 2026
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

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