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Alex17521 [72]
3 years ago
5

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

by 45% due to a friction force that is 25% of her weight. Use the work‑energy theorem to find the length of this rough patch.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer: 1.25 m

Explanation:

Given

initial velocity(v_i )=3 m/s

Final velocity(v_f)=0.55\times 3=1.65 m/s

Change in kinetic Energy =work done by Friction

change in Kinetic Energy=\frac{m}{2}\left ( v_i^2-v_f^2\right )

work done by friction=\mu mgL

\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L

3.135=0.25\times 9.8\times L

L=1.25

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
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Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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