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A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/m spring.


How far do they compress the spring before coming momentarily to rest? Ignore friction.

Respuesta :

Answer:

[tex]\Delta x=245\ mm[/tex]

Explanation:

Given:

  • mass of skier, [tex]m=70\ kg[/tex]
  • initial velocity of skier, [tex]u=4\ m.s^{-1}[/tex]
  • height of the hill, [tex]h=2\ m[/tex]
  • spring constant, [tex]k=2800\ N.m^{-1}[/tex]

final velocity of skier before coming in contact of spring:

Using eq. of motion:

[tex]v^2=u^2+gh[/tex]

[tex]v^2=4^2+9.8\times 2[/tex]

[tex]v=5.9666\ m.s^{-1}[/tex]

Now the time taken by the skier to reach down:

[tex]v=u+gt[/tex]

[tex]5.9666=4+9.8\ t[/tex]

[tex]t=0.2007\ s[/tex]

Now we calculate force using Newton's second law:

[tex]F=\frac{dp}{dt}[/tex]

[tex]F=\frac{m(v-u)}{t}[/tex]

[tex]F=\frac{70\times(5.9666-4)}{0.2007}[/tex]

[tex]F\approx686\ N[/tex]

∴Compression in spring before the skier momentarily comes to rest:

[tex]\Delta x=\frac{F}{k}[/tex]

[tex]\Delta x=\frac{686}{2800}[/tex]

[tex]\Delta x=0.245\ m[/tex]

[tex]\Delta x=245\ mm[/tex]

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