Question

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.

Answer 1

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

• mass of skier, $m=70\ kg$

• initial velocity of skier, $u=4\ m.s^{-1}$

• height of the hill, $h=2\ m$

• spring constant, $k=2800\ N.m^{-1}$

final velocity of skier before coming in contact of spring:

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

Now the time taken by the skier to reach down:

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

Now we calculate force using Newton's second law:

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

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

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$