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Chapter 6: Problem 61

A gymnast of mass \(52 \mathrm{kg}\) is jumping on a trampoline. She jumps so that her feet reach a maximum height of \(2.5 \mathrm{m}\) above the trampoline and, when she lands, her feet stretch the trampoline down \(75 \mathrm{cm} .\) How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]

Short Answer

Expert verified
Based on the given information about the gymnast and the trampoline, when the gymnast stands on the trampoline at rest, it stretches approximately 11.23 cm.

Step by step solution

01

1. Calculate the potential energy at the highest point of the jump

At the highest point of the jump, the gymnast has maximum potential energy with respect to the trampoline and no kinetic energy. The potential energy can be calculated using the formula: \(PE = mgh\), where \(PE\) is the potential energy, \(m\) is the mass of the gymnast, \(g\) is the acceleration due to gravity, and \(h\) is the maximum height above the trampoline. Given the mass (\(m = 52 kg\)) and height (\(h = 2.5 m\)), we can calculate the potential energy: \(PE = (52 kg)(9.81 \frac{m}{s^2})(2.5 m) = 1271.55 J\) (Joules).
02

2. Calculate the elastic potential energy when she lands

When the gymnast lands, her feet stretch the trampoline down 75 cm (\(h' = 0.75 m\)). This represents maximum elastic potential energy stored in the trampoline. Since the trampoline obeys Hooke's Law, the elastic potential energy can be calculated using the formula: \(EPE = \frac{1}{2}k(h')^2\), where \(EPE\) is the elastic potential energy, \(k\) is the spring constant of the trampoline, and \(h'\) is the distance the trampoline is stretched. At this point, energy conservation states that the potential energy at the highest point of the jump equals the elastic potential energy when she lands: \(PE = EPE \Rightarrow 1271.55 J = \frac{1}{2}k(0.75 m)^2\)
03

3. Calculate the spring constant (k)

Solve the equation from step 2 for the spring constant (k): \(1271.55 J = \frac{1}{2}k(0.56 m^2) \Rightarrow k=\frac{2(1271.55 J)}{0.56 m^2} = 4541.96 \frac{N}{m}\)
04

4. Calculate the elastic force when standing at rest

When the gymnast stands on the trampoline at rest, the gravitational force of her weight is equal to the elastic force exerted by the trampoline (reflecting that she's not accelerating). The gravitational force is equal to \(F_g = mg = (52 kg)(9.81 \frac{m}{s^2}) = 510.12 N\) (Newtons).
05

5. Calculate the stretching distance when standing at rest

Using Hooke's Law, we can express the elastic force as \(F_e = kh\), where \(F_e\) is the elastic force, \(k\) is the spring constant, and \(h\) is the stretching distance. Then, solve for the stretching distance when the gymnast stands on the trampoline at rest: \(510.12 N = (4541.96 \frac{N}{m})h \Rightarrow h = \frac{510.12 N}{4541.96 \frac{N}{m}} = 0.1123 m\) So, when the gymnast stands on the trampoline at rest, it stretches approximately 11.23 cm.

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