What hanging mass will stretch a 2.1-m-long, 0.43 mm - diameter steel wire by 1.5 mm ? the young'? s modulus of steel is 20×1010n/m2. express your answer using two significant figures?

The force (weight) due to the hanging mass is
F = mg
where
m = mass, kg
g = 9.8 m/s²

The cross sectional area of the wire is
A = (π/4)*(0.43×10⁻³ m)² = 1.4522×10⁻⁷ m²

The stress in the wire is
σ = F/A = (9.8m N)/(1.4522×10⁻⁷ m²) = 6.7484x10⁷m Pa

The length of the wire is 2.1 m and the extension is 1.5 mm. Therefore the induced strain is
ε = (1.5x10⁻³ m)/(2.1 m) = 7.1429x10⁻⁴

By definition, Young's modulus is E = σ/ε. Therefore
(6.7484x10⁷m Pa)/(7.1429x10⁻⁴) = 20x10¹⁰ Pa
9.4477x10¹⁰m = 20x10¹⁰
m = 2.117 kg 

2.12 kg (to two sig. figures)

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.