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3 years ago

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a woman weighing 500 N glides across some ice, starting her glide with a speed of 4.00 m/s. if the coefficient of friction betwe

en the skates and the ice is 0.115. how far does she go before coming to rest?

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

7.1 meters

Explanation:

We need to apply Newton's second law which explains the dynamics of objects under the effect of net forces.

The woman described in the question weighs 500N and has a speed of 4 m/s. If the net force acting on her was zero, she would continue to move at that speed forever (Newton's first law). But we know that there is friction and that force always goes against the movement. In the absence of a counterforce, the friction force is unbalanced and the woman will eventually stop.

The weight of the woman is

$W=m.g$

We can know her mass

$m=\frac{W}{g}=\frac{500}{9.8}=51.02\ Kg$

We know that the friction force is

$F_r=\mu N$

Where N is the normal force, which in this conditions is equal to the weight. So the net force is

$F_{net}=-\mu W=-0.115 (500Nw)=-57.5 Nw$

It's negative because it's oppossed to the movement, assumed to the right side. Since

$F_{net}=m.a$

$a=\frac{F_{net}}{m}=\frac{-57.5}{51.02}=-1.127\ m/sec^2$

From the formulas of cinematics we know:

$v_f^2=v_o^2+2ax$

Solving for x when $v_f=0$

$x=-\frac{v_o^2}{2a}=-\frac{4^2}{2(-1.127)}$

$x=7.1\ meters$

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