Question

Kate, a bungee jumper, wants to jump off the edge of a bridge thatspans a river below. Kate has a mass m, and the surface of thebridge is a height h above the water.The bungee cord, which haslength L when unstretched, will first straighten and then stretchas Kate falls.

Assume the following:
? The bungee cord behaves as an ideal spring once it begins tostretch, with spring constant k.
? Kate doesn't actually jump but simply steps off the edge of thebridge and falls straight downward.
? Kate's height is negligible compared to the length of the bungeecord. Hence, she can be treated as a point particle.

Use g for the magnitude of the acceleration due to gravity(don’t put numbers in - give both answers in terms of thealgebraic quantities in the statement).
(a) How far below the bridge will Kate eventually be hanging, onceshe stops oscillating and comes finally to rest? Assume that shedoesn't touch the water.
(b) If Kate just touches the surface of the river on her firstdownward trip (i.e. before the first bounce), what is the springconstant k?

Answer 1

Concepts and reason

The concept required to solve the given problem is law of conservation of momentum and Hooke’s law.

In the first part, calculate the distance to which Kate falls with the help of Hooke’s law.

In the second part, use the law of conservation of energy to calculate the spring constant.

Fundamentals

Newton’s Second Law: According, to Newton’s second law of motion the force applied is directly proportional to the acceleration of the object. Object with zero force has zero acceleration which further implies that the object moves with constant velocity.

The equation of the Newton’s second law is,

$\Sigma \vec F = m\vec a$

Here, $\Sigma \vec F$ is the net force on the object, $m$ is mass of the object, and $\vec a$ is the acceleration of the object.

The equilibrium condition for the force gives,

$\Sigma F = 0$

Here, $F$ is force.

Hooke’s Law: It states that the restoring force acting on a body is directly proportional to the amount of stretch.

$F = kx$

Here, $F$ is the force, $k$ is the spring and $x$ is the elongation.

Law of conservation of energy: It states that the total energy of an isolated system for a given frame of reference remains constant.

(A)

According, to Hooke’s law, the restoring force acting on a body is directly proportional to the amount of stretch.

$F = kx$

Here, $F$ is the force, $k$ is the spring and $x$ is the elongation.

The net force acting on the body is,

$\Sigma F = mg - kx$

Apply the equilibrium condition of force on Kate.

$\Sigma F = 0$

Substitute $mg - kx$ for $\Sigma F$ in equation $\Sigma F = 0$ and rearrange the equation for x.

$\begin{array}{c}\\mg - kx = 0\\\\x = \frac{{mg}}{k}\\\end{array}$

The distance below the bridge where Kate will be hanging is given by,

$d = x + L$

Substitute $\frac{{mg}}{k}$ for $x$in equation $d = x + L$.

$d = \frac{{mg}}{k} + L$

(B)

Apply the law of conservation of energy when Kate falls off from the bridge and touches the surface of river on her first downward trip.

$mgh = \frac{1}{2}k{x^2}$

Here, $m$ is the mass of Kate, $g$ is acceleration due to gravity, $h$is the height of bridge above water, $k$ is the spring constant and $x$ is the elongation of bungee cord.

The elongation of the spring will be equal to,

$x = h - L$

Here, $L$ is the length if bungee cord.

Substitute $h - L$ for $x$ in equation $mgh = \frac{1}{2}k{x^2}$.

$mgh = \frac{1}{2}k{\left( {h - L} \right)^2}$

Rearrange the above equation to determine the expression for the spring constant, $k$.

$k = \frac{{2mgh}}{{{{\left( {h - L} \right)}^2}}}$

Ans: Part A

The distance below the bridge where Kate will be hanging is $\frac{{mg}}{k} + L$.

Part B

The expression for spring constant is, $\frac{{2mgh}}{{{{\left( {h - L} \right)}^2}}}$.

Answer 2

a/ When she reaches the equilibrium, total force equalzero,Where x is stretched portion of the coreKate will be hanging below the bridge by a distance =L+x = L+mg/kb/ When she barely touches the water on the first downwardtrip her speed is zero and her distance from the bridge is h. Thecore stretches an amountequal to x=h-LHer initial potential energy (relative to the water surface)is mgh. When she touches the water and speed=0 the whole potentialenergy is converted tothe spring energy:E_sp = =mgh (where x=h-L)