Abungee cord can stretch, but it is never compressed. when thedistance between the two ends of the cord is less than its unstretched lengthl0, the cord folds and its tension is zero. for simplicity, neglect the cord’s ownweight and inertia as well as the air drag on the ball and the cord. a bungee cord has lengthl0 = 33 m when unstretched; when it’s stretched to l > l0, thecord’s tension obeys hooke’s law with "spring" constant 39 n/m. to test the cord’sreliability, one end is tied to a high bridge of height 94 m above the surface of a river) and the other end is tiedto a steel ball of mass75 kg. the ball is dropped off the bridge with zero initial speed. fortunately, the cord works and the ball stops in the air a fewmeters before it hits the water — and then the cord pulls itback up. the acceleration of gravity is 9.8 m/s2 . calculate theball’s height above the water’s surface at this lowest point of its trajectory. what is the tension of the bungee cord when the ballis at its lowest point?

a) 58.83 m b) 2262 N

Explanation:

The cord force constant K = 39 N/m, initial length of the cord L₀ = 33 m,

new length of the cord after the ball was just before the ball was pulled back L₁ = d + L₀ where d is the extension of the cord in meters

Weight of the ball = m g where m is the mass of the ball and g is acceleration due to gravity in m/s² = 75 × 9.8 = 735 N

loss of potential energy = equal to gain in elastic potential energy by the cord

loss in potential energy = mgh = 735 × ( d + L₀)

and the elastic potential gain by the cord = 1/2 × K ×d²

735(d + 33) = 0.5×39×d²

735(d + 33) = 19.5 d²

735 d + 24255 = 19.5 d²

rearrange the equation

-19.5 d² + 735 d + 24255 = 0

multiply both side of the equation with -1

19.5 d²  - 735 d - 24255 = 0

solve the equation using  quadratic formula

-b ± √(b² - 4ac) / 2a

where a = 19.5, b = -735, c = -24255

substitute this values into the formula

- (-735) ±√((-735²) - (4×19.5×-24255)) / (2×19.5)

735±√(540225 + 1891890) / (39)

735 ± √(2432115) / 39

735 ± 1559.5 / 39

(735+ 1559.5) / 39 or (735 - 1559.5) / 39

58.83 m or -21.1 m

the new length = 33 m+ 58.83 m = 91.83

                        or

the new length = 33 m + (-21) = 12

using the positive value

the ball's height above the water = 94 - 91.83 = 2.17 m

b) Tension in the cord = Kd using hooke's law

T = 39 × 58 = 2262 N

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

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