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mihalych1998 [28]
3 years ago
8

A constant force of friction 50N is acting on a body of mass 200 Kg moving initially with a speed of 15 m/s. How long does the b

ody take to stop? What distance will it cover before coming to rest?
Physics
1 answer:
Zina [86]3 years ago
8 0

Answer:

F=-50N

M=200kg

U=15m/s

F=Ma

a=F/M=-50/200=-0.25

V^2-U^2=2aS

0-(15)^2=2(-0.25)S

S=-225/-0.5=450m

V=U+at

0=15-0.25t

t=-15/-0.25=60s

Explanation:

Hope this helps, let me know if you have any questions!

Have a great day.

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stealth61 [152]

Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

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The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

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=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

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Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

=>      S = 24.86 ft        

                 

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Answer:

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