Question

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is 9.6 m/s as y = 0. What is her kinetic energy at this point? (c) What is her kinetic energy at maximum height? 0 Correct: Your answer is correct. J What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. e) Write the general equation for energy conservation and solve for the velocity at half the maximum height.

Answer 1

The interacting objects are the athlete and the trampoline. The athlete's initial kinetic energy and kinetic energy at the height of 9.6 m/s can be calculated using the mass and speed. The gravitational potential energy at maximum height can be calculated using the mass, acceleration due to gravity, and height. The maximum height can be determined by solving the energy conservation equation. The velocity at half maximum height can be found by solving another energy conservation equation.

Step-by-step explanation:

(a) In this problem, the interacting objects are the athlete and the trampoline. The athlete interacts with the trampoline through contact forces. The athlete exerts a downward force on the trampoline, and the trampoline exerts an equal and opposite upward force on the athlete, according to Newton's third law of motion.

(b) If we select the height at which the athlete's speed is 9.6 m/s as y = 0, then her kinetic energy at this point would be given by the formula KE = 0.5 * m * v^2, where m is the mass of the athlete and v is her speed. Plugging in the values, we get KE = 0.5 * 62.0 kg * (9.6 m/s)^2 = 2884.16 J.

(c) At maximum height, the athlete's kinetic energy would be zero because she has come to a stop. Therefore, her kinetic energy at maximum height is 0 J. The gravitational potential energy associated with the athlete can be calculated using the formula PE = m * g * h, where m is the mass of the athlete, g is the acceleration due to gravity, and h is the height. Plugging in the values, we get PE = 62.0 kg * 9.8 m/s^2 * h.

(d) The general equation for energy conservation in this case is KE + PE = PE_max, where KE is the kinetic energy, PE is the gravitational potential energy, and PE_max is the maximum potential energy. Solving for the maximum height, we get h = (KE + PE_max) / (m * g).

(e) The general equation for energy conservation is KE + PE = PE_max. At half maximum height, the kinetic energy would be half of the maximum potential energy. Therefore, we can write the equation as 0.5 * KE_max + PE_half = PE_max. Solving for the velocity at half maximum height, we get v_half = sqrt((2 * (PE_max - PE_half)) / m).

Answer 2

2856.96 J

0

0

`(1)/(2)mv_i^2+mgh_i=(1)/(2)mv_f^2+mgh_f`

6.78822 m/s

Step-by-step explanation:

`v_i` = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

`K=(1)/(2)mv^2\\\Rightarrow K=(1)/(2)* 62* 9.6^2\\\Rightarrow K=2856.96\ J`

At height y = 0 the potential energy is 0 as

`P=mgy\\\Rightarrow P=mg0=0`

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

`K_i+P_i=K_f+P_f\\\Rightarrow (1)/(2)mv_i^2+mgh_i=(1)/(2)mv_f^2+mgh_f`

Half of maximum height

`\\\Rightarrow mgh_i+(1)/(2)mv_f^2=mg(h_i)/(2)+(1)/(2)mv^2\\\Rightarrow gh_i=g(h_i)/(2)+(1)/(2)v^2\\\Rightarrow g(h_i)/(2)=(1)/(2)v^2\\\Rightarrow v=√(gh)`

`h_i=(v_i^2)/(2g)`

`v=√(gh)\\\Rightarrow v=\sqrt{g* (v_i^2)/(2g)}\\\Rightarrow v=\sqrt{(v_i^2)/(2)}\\\Rightarrow v=\sqrt{(9.6^2)/(2)}\\\Rightarrow v=6.78822\ m/s`

The velocity of the athlete at half the maximum height is 6.78822 m/s