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Phantasy [73]
3 years ago
14

A luge and its rider, with a total mass of 85 kg, emerge from adownhill track onto a horizontal straight track with an initial s

peedof 37 m/s. if a force slows them to a stop at a constant rate of 2.0m/s2, (a) what magnitude f is required for the force, (b) what distanced do they travel while slowing, and (c) what work w is doneon them by the force? what are (d) f, (e) d, and (f) w if they, instead,slow at 4.0 m/s2?
Physics
1 answer:
Verdich [7]3 years ago
7 0

<span>(1)   </span>Through the Second Law of motion, the equation  for Force is:

                                 F = m x a

Where m is mass and a is acceleration (deceleration)

 

<span>(2)   </span>Distance is calculated through the equation,

                             D = Vi^2 / 2a

Where Vi is initial velocity

<span>(3)   </span>Work is calculated through the equation,

                          W = F x D

 

Substituting the known values,

Part A:

<span>(1)   </span>  F = (85 kg)(2 m/s^2) = 170 N

<span>(2)   </span>  D = (37 m/s)^2 / (2)(2 m/s^2)  = 9.25 m

<span>(3)   </span>  W = (170 N)(9.25 m) = 1572.5 J

 

Part B:

<span>(1)   </span> F = (85 kg)(4 m/s^2) = 340 N

<span>(2)   </span>D = (37 m/s)^2 / (2)(4 m/s^2) = 4.625 m

<span>(3)   </span><span> W = (340 N)(4.625 m) = 1572.5 J</span>

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A child of mass 46.2 kg sits on the edge of a merry-go-round with radius 1.9 m and moment of inertia 130.09 kg m2 . The merrygo-
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Answer:

The angular velocity is w_f = 4.503 \  rad/s

Explanation:

From the question we are told that

   The mass of the child is  m_c  =  46.2 \ kg

    The radius of the merry go round is  r =  1.9 \ m

     The moment of inertia of the merry go round is I_m =  130.09 \  kg \cdot  m^2

      The angular velocity of the merry-go round is  w =  2.4 \ rad/s

       The position of the child from the center of the merry-go-round is  x = 0.779 \ m

According to the law of angular momentum conservation

    The initial angular momentum  =  final  angular momentum

So  

       L_i  =  L_f

=>     I_i w_i  =  I_fw_f

Now   I_i is the initial moment of inertia of the system which is mathematically represented as

          I_i  = I_m + I_{b_1}

Where  I_{b_i} is the initial moment of inertia of the boy which is mathematically evaluated as

      I_{b_i} =  m_c * r

substituting values

      I_{b_i} =  46.2 *  1.9^2

      I_{b_i} =  166.8 \ kg \cdot m^2

Thus

   I_i  =130.09 + 166.8        

   I_i  = 296.9 \ kg \cdot m^2      

Thus  

     I_i * w_i  =L_i=  296.9 * 2.4

       L_i  = 712.5 \ kg \cdot m^2/s

Now  

     I_f =  I_m  + I_{b_f }

Where  I_{b_f} is the final  moment of inertia of the boy which is mathematically evaluated as

         I_{b_f} =  m_c * x

substituting values

         I_{b_f} =  46.2 * 0.779^2

         I_{b_f} =  28.03  kg \cdot m^2

Thus

      I_f =  130.09 + 28.03

      I_f =  158.12 \ kg \ m^2

Thus

     L_f  =  158.12 * w_f

Hence

      712.5  =  158.12 * w_f

       w_f = 4.503 \  rad/s

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Answer:

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