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A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.06 m/s. Her hands are in contact with the wall for 1.07 s. Ignore friction and wind resistance. Find the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her). Note that this force has direction, which you should indicate with the sign of your answer.

Respuesta :

cjrodriguez89

Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       [tex]F = \frac{\Delta p}{\Delta t}[/tex]

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       [tex]F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N[/tex]

       ⇒ Fsk = 47.6 N (normal to the wall)

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