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Grade: 11
        
Determine the horizontal force required to displace a mass of 0.03 KG suspended by a string until the string makes an angle 30 degree with the vertical. Please solve it by method of resolution of vectors. 
5 months ago

Answers : (2)

sree lakshmi
39 Points
							
Given that mass =0.03kg
Length of string =l
Using the law of conservation of mechanical energy
Work done by the horizontal force= gain in PE
ie F×l sin30 =mgl(1-cos30)
F=mg(1-cos30)/sin30
F=0.03×10×(1-√3/2)÷1/2
=0.3×(2-√3)
=0.0804 N

   
5 months ago
Khimraj
3008 Points
							
Given,
m = 0.03 kg
Using the law of conservation of mechanical energy,
Work done by the horizontal force = Gain in P. E. of mass m
or, F × 1 sin 30° = mg × ( 1 - 1 cos 30°)
or, F = mg × (1 -1 cos 30°)/1 sin 30°
= 0.03 g × (1 - √3/2)/(1/2)
= 0.03 g × (2 - √3/2)/(1/2)
= 0.03 g × (2 - √3)
= 0.03 g × 9.8 × (2 - √3)
= 0.079 N Answer
 
5 months ago
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