Can someone help me through is problem?? Thanks! A bungee jumper leaps from a br

Question

Can someone help me through is problem?? Thanks!

A bungee jumper leaps from a bridge and undergoes a series of oscillations. Assume g = 9.78 m/s2.

a) If a 60.0 kg jumper uses a bungee cord that has an unstretched length of 33.0 m and she jumps from a height of 50.0 m above a river, coming to rest just a few centimeters above the water surface on the first downward descent, what is the period of the oscillations? Assume the bungee cord follows Hooke's law.

b) The next jumper in line has a mass of 80.0 kg. Should he jump using the same cord? Explain.

Explanation / Answer

(a) By the time the cord is fully stretched, all the jumper'sgravitational potential energy has been converted into elasticpotential energy in the cord.

0.5 k x2 = m g y

0.5 k (50.0 - 33.0)2 = 60.0 * 9.78 * 50.0

? k = 203 N/m

The jumper's velocity after 33.0 m of free fall is v2 =2 g y = 2 * 9.78 * 33.0

? v = 25.4 m/s

At the equilibrium position, m g = k x

60.0 * 9.78 = 203 x

x = 2.89 m

Therefore, the equilibrium position is (33.0 + 2.89) m below thebridge.

Fav = ma = m ?v / ?t
Fav ?t = m ?v
Since the cord's restoring force varies from 0 (when x = 0) to kx,its average value is (0 + kx)/2 = 0.5 k x
0.5 k x ?t = m v
0.5 * 203 (50.0 - 33.0) ?t = 60.0 * 25.4
? ?t = 0.883 s (this is the time the jumper takes toaccelerate from 25.4 m/s to 0 m/s at the bottom of the jump)

? At time t = 0.833 s, x = -2.89 m (taking downward as thepositive direction)

x = A cos (? t) = (50.0 - 33.0 - 2.89) cos (? t) = 14.1cos (? t)

-2.89 = 14.1 cos (0.833 ?)

? ? = 2.13 rad/s = 2 ? / T

? T (period) = 2 ? / ? = 2.94 s

(b) At the bottom of the jump, the jumper has lost all of hismechanical energy (gravitational potential and kinetic) in the formof elastic potential energy in the spring.

0.5 m v2 + m g x = 0.5 k x2

0.5 * 80.0 * 25.42 + 80.0 * 9.78 * x = 0.5 * 203 *x2

101.5 x2 - 782.4 x - 25806.4 = 0

? x = 20.3 m (ignore the negative solution because x ispositive downward)

? The total distance traveled by the 80.0 kg jumper is 33.0 +20.3 = 53.3 m which is larger than the bridge's altitude.Therefore, the 80.0 kg person should not jump using the samecord.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.