A 70 kg bicyclist rides his 9.8 kg bicycle with a speed
1 answer:
<h2>Answer:</h2>
<em>Hello, </em>
<h3><u>QUESTION)</u></h3><em>✔ We have Ek = 1/2m x v² </em>
- Ek = 1/2 x 79.8 x 16²
- Ek = 10 214.4 J
In order to come to a complete stop, the cyclist must convert all his kinetic energy into thermal energy. Given that the braking force opposes movement, the work is therefore resistant, i.e. W = -10 214.4 J.
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Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) =
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
