Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.18 ✕ 10⁴ N/m.

A) What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 78.0 kg?

B) How much would this rope stretch (in cm) to break the climber's fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy.

C) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

frequency (in Hz)

stretch length (in cm)

Answer 1

a.)

Frequency rope is given by,

f = (1/(2*pi))*sqrt(k/m)

given, k = force constant of rope = 1.18*10^4 N/m

m = mass = 78.0 kg

So, f = (1/(2*pi))*sqrt[(1.18*10^4)/78.0]

f = 1.96 Hz

b.)

By energy conservation,

KEi + PEi + SEi = KEf + PEf + SEf

here, KEi = initial kinetic energy = 0

PEi = Initial potential energy = m*g*di

SEi = initially energy stored in spring = 0 (Because initially it is not stretched)

KEf = Final kinetic energy = 0

PEf = Initial potential energy = m*g*df

SEf = initially energy stored in spring = 0.5*k*x^2 (Let it stretched length 'x')

given, di = 0 (at top)

df = total distance he falls = -(h+x) m

h = 2.00 m

k = 1.18*10^4 N/m

So, 0 + 0 + 0 = m*g*df + 0 + 0.5*k*x^2

0.5*k*x^2 = m*g*(h+x)

0.5*(1.18*10^4)*x^2 = 78.0*9.81*(2 + x)

0.5*1.18*10^4*x^2 - 78.0*9.81*x - 78.0*9.81*2 = 0

By solving quadratic equation,

x = 0.578 m

x = 57.8 cm

C.)

When the length of nylon rope become twice,

in this case value of Force constant of rope = k' = k1*k2/(k1+k2)

k' = k/2 = (1.18*10^4)/2

k' = 0.59*10^4 N/m

then, new frequency will be,

f' = (1/(2*pi))*sqrt(k'/m)

f' = (1/(2*pi))*sqrt[(0.59*10^4)/86.0]

f' = 1.32 Hz

now, By energy conservation,

KEi + PEi + SEi = KEf + PEf + SEf

here, KEi = initial kinetic energy = 0

PEi = Initial potential energy = m*g*di

SEi = initially energy stored in spring = 0 (Because initially it is not stretched)

KEf = Final kinetic energy = 0

PEf = Initial potential energy = m*g*df

SEf = initially energy stored in spring = 0.5*k'*x^2 (Let it stretched length 'x')

0 + 0 + 0 = m*g*df + 0 + 0.5*k'*x^2

0.5*k'*x^2 = m*g*(h+x)

0.5*(0.59*10^4)*x^2 = 78.0*9.81*(2 + x)

(0.5*0.59*10^4)*x^2 - 78.0*9.81*x - 78.0*9.81*2 = 0

By solving quadratic equation,

x = 0.861 m

x = 86.1 cm

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