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faltersainse [42]
3 years ago
7

The stair stepper is a novel exercise machine that attempts to reproduce the work done against gravity by walking up stairs. Wit

h each step, Brad (of mass 60 kg) simulates stepping up a vertical distance of 0.2 m with this machine. If Brad exercises for 13 min per day with a stair stepper at a frequency of 50 steps per minute, what total work he does he do each day
Physics
1 answer:
Delicious77 [7]3 years ago
7 0

Answer:

W = 76518 J = 76.518 KJ

Explanation:

First, we will find the total number of steps taken by Brad, as follows:

No.\ of\ steps = n = (13\ min)(50\ steps/min) = 650\ steps\\

Now, we will calculate the height covered while covering these number of steps:

Height\ Covered = H = (n)(Height\ for\ 1\ step)\\\\H = (0.2\ m/step)(650\ steps)\\H = 130\ m

Now, we will calculate the amount of work done while covering this distance:

Work = W = mgH = \\W = (60\ kg)(9.81\ m/s^2)(130\ m)

<u>W = 76518 J = 76.518 KJ</u>

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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

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Explanation:

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d = Diameter

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v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

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Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

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There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

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