Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm.

a) What average force is exerted on the nail?
b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long?
c) What pressure is created on the 1.00-mm-diameter tip of the nail?

Answer 1

Answer:

(a) 20089.29 N

(b) 0.00123 m

(c) 2.511×10¹⁰ N/m²

Explanation:

(a)  kinetic Energy of the hammer = Work done on the nail. ( assuming no energy is lost to heat)

Ek = 1/2mv²........................... Equation 1

Work done = F×d................ Equation 2

Therefore,

F×d = 1/2mv²

making F the subject of the equation above,

F = 1/2mv²/d ....................... Equation 3

Where F = Average force exerted on the nail, m = mass of the hammer, v = velocity d = distance.

Given: m = 0.5 kg, d = 2.80 mm = 0.0028 m, v = 15.0 m/s.

Substituting into equation 3,

F = 1/2(0.5)(15)²/(0.0028)

F = 20089.29 N.

(b) Young's modulus = stress/Strain

Where stress = F/A

and     Strain = ΔL/L

Therefore,

γ = F(L)/AΔL

ΔL = FL/Aγ.............................. Equation 4

Where γ = young's modulus of steel, A = cross sectional area of the nail, L = length of the nail, F = average force exerted on the nail.

Given: F = 20089.29 N, L = 6.00 cm = 0.06 m, A = πd²/4

Where d = diameter = 2.5 mm = 0.0025 m, π = 3.143

A = 3.143(0.0025)²/4 = 0.0000049 m².

Constant: γ = 200 GN/m² = 200×10⁹ N/m².

Substituting into equation 4

ΔL =20089.29(0.06)/(200×10⁹×0.0000049)

ΔL = 1205.356/980000

ΔL = 1205.356/980000

ΔL = 0.00123 m

(c) Pressure = Force/Area

P = F/A................. Equation 4

Where P = pressure, F = Force, A = area.

Given: F = 20089.29 N, A = πd²/4

Where d = diameter = 1 mm = 0.001 m, π = 3.143

A = 3.143(0.001)²/4 = 0.0000008 m².

Substituting into equation 4

P = 20089.29/0.0000008

P = 2.511×10¹⁰ N/m²

Answer 2

(a) The average force exerted on the nail is 20,089.3 N.

(b) The compression of the nail is 1.2 mm.

(c) The pressure created on 1 mm diameter tip of the nail is $2.56 \times 10^{10} \ N/m^2$

The given parameters;

• mass of the hammer, m = 0.5 kg
• initial velocity of the hammer, v = 15 m/s
• distance of the hammer, d = 2.8 mm = 0.0028 m
• diameter of the nail, D = 2.5 mm = 0.0025 m
• length of the nail, l = 6 cm = 0.06 m

(a) Apply the principle of work-energy theorem to determine the average force exerted on the nail;

$W = K.E\\\\Fd = \frac{1}{2} mv^2\\\\F = \frac{mv^2}{2d} \\\\F = \frac{0.5\times 15^2}{2 \times 0.0028} \\\\F = 20,089.3 \ N$

(b) The compression of the nail is determined by applying Young's modulus equation;

$E = \frac{Fl}{A \Delta x} \\\\\Delta x = \frac{Fl}{AE}$

where;

• A is the area of the nail
• E is Young's modulus of elasticity for steel = 200 x 10⁹ N/m²

$A = \frac{\pi D^2}{4} \\\\A = \frac{\pi \times 0.0025^2}{4} \\\\A = 4.91 \times 10^{-6} \ m^2$

$\Delta x = \frac{Fl}{AE} \\\\\Delta x = \frac{(20,089.3) \times (0.06)}{(4.91 \times 10^{-6}) \times (200 \times 10^9)} \\\\\Delta x = 0.0012 \ m\\\\\Delta x = 1.2 \ mm$

(c) The pressure created on 1 mm diameter tip of the nail;

$A = \frac{\pi D^2 }{4} = \frac{\pi \times 0.001^2}{4} = 7.86 \times 10^{-7} \ m^2\\\\P = \frac{F}{A} \\\\P = \frac{20,089.3}{7.86 \times 10^{-7}} \\\\P = 2.56 \times 10^{10} \ N/m^2$

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