Answer to Question #269780 in Mechanics | Relativity for efe

Question

Two rock climbers, Bill and Karen, use safety ropes of similar length. Karen's rope is more elastic, called a dynamic rope by climbers. Bill has a static rope, not recommended for safety purposes in pro climbing.

a)Karen falls freely about 2.1 m

m and then the rope stops her over a distance of 1.1 m

m . Estimate how large a force (assume constant) she will feel from the rope. (Express the result in multiples of her weight.)

b)In a similar fall, Bill's rope stretches by only 45 cm

cm . How many times his weight will the rope pull on him?

c)Which climber is more likely to be hurt?

Answer 1

$a) \text{for \ Karen:}$

$P_k= m_kg-\text{weight Karen}$

$\text {the beginning of the fall:}$

$h_0= 0$

$v_0 = 0$

$E =EP = 0$

$\text{end of fall:}$

$h = -2.1m$

$\Delta x=-1.1m$

$h_1 = h+\Delta x=-3.2m$

$v_1=0,\text{hense }EK =0$

$E = EP_k+EP_r =0$

$EP_k-\text{Karen's potential energy}$

$EP_r\text{ -potential energy of the rope}$

$EP_k = m_kgh_1=-3.2m_kg$

$EP_r = -\frac{F_r \Delta x}{2}= 0.55F_r$

$-3.2m_kg+0.55F_r= 0$

$F_r = 5.82m_kg=5.82P_k$

$\text{Answer:}$

$\text{The force exerted by the rope is }5.82\text{ times the weight}$

$b)\text{ \text{for \ Bill:}}$

$\text{end of fall:}$

$h= -2.1m$

$\Delta x'=45cm =0.45m$

$h'_1 = h+\Delta x'=-2.55m$

$EP_b = m_bgh'_1=-2.55m_bg$

$EP'_r = -\frac{F'_r \Delta x'}{2}= 0.225F_r$

$-2.55m_bg+0.225F'_r= 0$

$F'_r = 11.33m_bg=11.33P_b$

$\text{Answer:}$

$\text{The force exerted by the rope is }11.33\text{ times the weight}$

$c)F_r =5.82P_k$

$F'_r =11.33P_b$

$a_k = 5.82g$

$a_b=11.33$

$\text{Bill's acceleration is greater than Karen's,}\newline \text{getting injured is also greater.}$

$\text{Answer:Bill is more likely to get hurt}$